Master
Mathematics AA
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[Lessons] to bring out the best in every student
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1.1 | Sequences and Series
What are sequences and series?
Although commonly mistaken for the same thing, sequences and series are 2 related, but different concepts. Both deal with patterns in lists of numbers, but the former describes the list itself, whilst the latter describes the sum of the list itself. In simple terms; sequences are the list of numbers, while series are the summation (addition) of numbers in the list.
In the IB, sequences can be given names in the form of \(u_n\), where \(u\) can be any letter or symbol, which just denotes the sequence (list of numbers), so that it's easy and simple to refer to. The \(n\) basically denotes that we are referring to the entire sequence, not just an arbitrary number. If \(n\) is replaced by a number, it refers to that term in the sequence, starting with 1. For instance, if \(n=1\), we are looking at the first term in the sequence.
For example, take the sequence \(u_n = 1, 3, 5, 7, 9, …\). When referring to this sequence, one would say \(u_n\), instead reciting all the numbers within the sequence. In this case, to refer to the first term in the sequence (\(n = 1\)), we would say \(u_1\). For the sequence \(u_n\), \(u_1 = 1\). In other words, the first term in the sequence \(u_n\) is 1. In the same way, \(u_2\) (the second term) is 3.
What if we want to look at the sum, or the series of the sequence \(u_n\)? To denote a series, we use the capital ‘S’, which stands for “sum”, with the number of terms added represented by \(n\), making for \(S_n\). If this sounds complicated, in practice it's really easy, don’t worry!
In essence, all \(S_n\) means is the “sum of \(n\) terms” of a sequence. Mathematically, \(S_n = u_1 + u_2 + u_3 + … + u_n\). Looking specifically at the sequence \(u_n\):
\[u_n = 1, 3, 5, 7, 9, …\]
\begin{align}S_3 &= u_1 + u_2 + u_3\\S_3 &= 1 + 3 + 5\\& \rightarrow \boxed{S_3 = 9} \\\end{align}
The common difference
There are 2 kinds of sequences, one of which is arithmetic. You may have noticed that the above sequence, \(u_n\), propagated by units of 2; 1 went to 3, 3 went to 5, and 5 went to 7, and so on. Essentially, 2 was being added to each consecutive term as the sequence went on.
This is an example of an arithmetic sequence. The terms propagate through addition (or subtraction).
Taking another sequence, \(a_n = 0, 5, 10, 15, 20, 25, …\), it can be noticed that the terms are being propagated by adding units of 5 this time, as 0 goes to 5, and 5 goes to 10. Although \(a_n\) may not seem like an arithmetic sequence, because it is similar to the 5 times-table, it is still arithmetic, as \(a_1 = 0\), and anything multiplied by 0 is still 0— there is no other possibility of propagation other than by adding 5 each time.
This amount by which the terms propagate in an arithmetic sequence is called the common difference, \(d\). In the case of \(a_n\), \(d = 5\).
To find the common difference, you can use the following equation:
\[d = {u}_{n+1} - u_n \text{ or } d = u_n - u_{n-1}\]
Basically, the common difference is any term in the arithmetic sequence subtracted by the previous term in the same sequence.
One can use this fact to verify whether the common difference is correct; using 3 terms and looking at the common differences between them:
\[a_n = 0, 5, 10, 15, 20, 25, …\]
\[a_1 = 0, a_2 = 5, a_3 = 10\]
\begin{align}d \Rightarrow {u}_{n+1} - u_n &= u_n - u_{n-1}\\a_3 - a_2 &= a_2 - a_1\\10 - 5 &= 5 - 0\\5 &= 5 \rightarrow \boxed{\text{True}} \\\end{align}
As both sides of the equation equal each other (true statement), the common difference is indeed verified to be 5 for \(a_n\). Keep in mind that you must keep the LHS and RHS seperate when attempting to verify the common difference!
The arithmetic \(n\)th term
Just by knowing the first term \(u_1\) and the common difference \(d\) of a sequence, any term in the arithmetic sequence, or a summation through a series can be found for that sequence. This is through the given formulas in the data booklet:
\[u_n = u_1 + (n-1)d\]
Through plugging in the values for \(u_1\) and \(d\), any term can be found by plugging in for the desired term in \(n\).
The same goes for the series, where plugging in the values for \(u_1\) and \(d\), any sum of terms up to the \(n\)th term can be determined by plugging in for \(n\):
\[S_n = {n \over 2}(2{u}_{1}+(n-1)d) \text{ or } S_n = {n \over 2}(u_1+u_n)\]
The common ratio
The other kind of sequence is geometric. Unlike arithmetic sequences, geometric sequences propagate by multiplication (or division).
Taking the sequence \(b_n = 1, 3, 9, 27, 81, …\), it can be noticed that the terms are being propagated by multiplying units of 3 this time, as 1 goes to 3, and 3 goes to 9. Clearly this is not an arithmetic sequence, as when one would try to find a common difference, it would lead to a false statement:
\[b_n = 1, 3, 9, 27, 81, …\]
\[b_2 = 3, b_3 = 9, b_4 = 27\]
\begin{align}d \Rightarrow {u}_{n+1} - u_n &= u_n - u_{n-1}\\b_4 - b_3 &= b_3 - b_2\\27 - 9 &= 9 - 3\\18 &= 6 \rightarrow \boxed{\text{False}} \\\end{align}
As both sides of the equation do not equal each other, a false statement is produced, meaning that the sequence \(b_n\) cannot be arithmetic.
This amount by which the terms propagate in a geometric sequence is called the common ratio, \(r\). In the case of \(b_n\), \(r = 3\).
To find the common ratio, you can use the following equation:
\[r = {{u}_{n+1} \over u_n} \text{ or } r = {u_n \over u_{n-1}}\]
Basically, the common ratio is any term in the geometric sequence divided by the previous term in the same sequence.
One can use this fact to verify whether the common ratio is correct; using 3 terms and looking at the common ratios between them:
\[b_n = 1, 3, 9, 27, 81, …\]
\[b_1 = 1, b_2 = 3, b_3 = 9\]
\begin{align}r \Rightarrow {{u}_{n+1} \over u_n} &= {u_n \over u_{n-1}}\\{u_3 \over u_2} &= {u_2 \over u_1}\\{9 \over 3} &= {3 \over 1}\\3 &= 3 \rightarrow \boxed{\text{True}} \\\end{align}
As both sides of the equation equal each other (true statement), the common ratio is indeed verified to be 3 for \(b_n\). Keep in mind that you must keep the LHS and RHS seperate when attempting to verify the common ratio!
The geometric \(n\)th term
In the same way as arithmetic sequences, the same goes for geometric: just by knowing the first term \(u_1\) and the common ratio \(r\) of a sequence, any term in the geometric sequence, or a summation through a series can be found for that sequence. This is through the given formulas in the data booklet:
\[u_n = u_1 \times {r}^{n-1}\]
Through plugging in the values for \(u_1\) and \(r\), any term can be found by plugging in for the desired term in \(n\).
The same goes for the series, where plugging in the values for \(u_1\) and \(r\), any sum of terms up to the \(n\)th term can be determined by plugging in for \(n\):
\[S_n = {{u}_{1}({r}^{n}-1) \over r-1} \text{ or } S_n = {{u}_{1}(1-{r}^{n}) \over 1-r}\]
Infinite geometric series (summation of infinite sequence)
Some geometric sequences are special, in that their summations (series) converge to a certain number when approaching infinity: \(n \rightarrow \infty\). To figure out whether the sum approaches a value (converges), or approaches positive or negative infinity (diverges), it is important to find the common ratio, and plug it into the following equation:
\[{S}_{\infty} = {u_1 \over 1-r}, |r|<1\]
As for why this equation only works for \(|r|<1\) (meaning that \(-1<r<1)\)), is because if \(r\) is not within this range, the equation is bound to diverge; approach infinity, instead of approaching a value.
To better understand this concept, we can take the example of the sequence:
\[c_n = 1, -{1 \over 3}, {1 \over 9}, -{1 \over 27}, {1 \over 81}, …\]
In this case, it can be identified that \(c_n\) can be rewritten in powers of 3:
\begin{align}c_n &= 1, -{1 \over 3}, {1 \over 9}, -{1 \over 27}, {1 \over 81}, …\\c_n &= 1, -{1 \over 3}, {1 \over {3}^{2}}, -{1 \over {3}^{3}}, {1 \over {3}^{4}}, …\\\end{align}
As previously stated, the first step is to find the common ratio, \(r\) of this sequence:
\[c_n = 1, -{1 \over 3}, {1 \over {3}^{2}}, -{1 \over {3}^{3}}, {1 \over {3}^{4}}, …\]
\[c_1 = 1, c_2 = -{1 \over 3}, c_3 = {1 \over {3}^{2}}\]
\begin{align}r \Rightarrow {{u}_{n+1} \over u_n} &= {u_n \over {u}_{n-1}}\\{c_3 \over c_2} &= {c_2 \over c_1}\\{{1 \over {3}^{2}} \over -{1 \over 3}} &= {-{1 \over 3} \over 1}\\{-3 \over {3}^{2}} &= -{1 \over 3}\\-{1 \over 3} &= -{1 \over 3}\\\end{align}
Thus, we have identified that because the equation is true, the sequence is indeed a geometric sequence, with a common ratio of \(r = -{1 \over 3}\). This also satisfies the condition that \(|r|<1\), because \(-1<-{1 \over 3}<1)\), meaning that this sequence converges to some constant value.
In order to find this value, we can substitute the values for \(c_1\) and \(r\) into the aforementioned equation for the sum of an infinite geometric sequence:
\begin{align}{S}_{\infty} &= {u_1 \over 1-r}\\{S}_{\infty} &= {c_1 \over 1-r}\\{S}_{\infty} &= {1 \over 1-(-{1 \over 3})}\\{S}_{\infty} &= \boxed{{3 \over 4}}\\\end{align}
This means that the summation of the sequence \(c_n\) approaches \({3 \over 4}\).
To visually understand this phenomenon, the graph of the series can be visualized:
[Questions] to train for any occasion
A molecule travels in a straight line. The velocity, \(v\) \({\text{ms}}^{-1}\) of the molecule at time \(t\) seconds is given by \(v(t) = t\cos(t) + 5\), where \(0 \leq t \leq 15\).
The following diagram displays the graph of \(v\):
(a) Find the smallest value of \({t}\) for which the molecule is at rest.
(b) Find the total distance traveled by the molecule.
(c) Find the acceleration of the molecule when \(t\) = 11.
(a) Find the smallest value of \(t\) for which the molecule is at rest.
Recognize that velocity \(v\) = 0 at rest.
[1]
Use G.D.C. to graph \(v(t) = t\cos(t) + 5\), and finding first x-intercept (because \(v = 0\) @ x-intercept)
\[t=\boxed{8.48 \text{ s}}\]
[1]
(b) Find the total distance traveled by the molecule.
Recognize that distance is area under curve of velocity v/s time graph.
Hence using G.D.C. :
\begin{align} d & = \int_{0}^{8.484}{t\cos(t) + 5\ dt} + \left| \int_{8.484}^{10.499}{t\cos(t) + 5}\text{dt} \right| + \int_{10.499}^{14.489}{t\cos(t) + 5\ dt} + \left| \int_{14.489}^{15}{t\cos(t) + 5\ dt} \right|\\ d & = \boxed{98.2 \text{ m}} \end{align}
[2]
(c) Find the acceleration of the molecule when \(t\) = 11.
Recognize that acceleration, \(a\), is given by \(\frac{\text{dv}}{\text{dt}}\) or \({v}^{'}\). Hence:
\begin{align*}
v( t) & = t\cos( t) +5\\
v'\ ( t) & = -t\sin( t) +\cos( t) = a( t)\\
a( 11) & = -11\sin( 11) +\cos( 11)\\
a( 11) & = \boxed{-1.12 \text{ ms}}
\end{align*}
[2]
Every Christmas (\({24}^{th}\) December) since 2003, Tanaka invested €\(d\) in a savings account to purchase a present one day for her younger brother, also born on Christmas day, 2003.
She continued to deposit €\(d\) on the \({24}^{th}\) of every month from then.
With a fixed rate of 0.32% interest per month, the interest was calculated on the last day of each month and subsequently added onto the account.
Let €\(D_{n}\)be the total amount in Tanaka's account on the last day of the \({n}^{th}\) month immediately after the interest has been added.
(a) Find an expression for \(D_{1}\) and show that \(D_{2} = {1.0032}^{2}d + 1.0032d\).
(b) (i) Find a similar expression for \(D_{3}\) and \(D_{4}\).
(ii) Hence, show that the amount in Tanaka's account the day before her brother's \({7}^{th}\) birthday is given by \(313.5({1.0032}^{83} - 1)d\).
(c) Write down an expression for \(D_{n}\) in terms of \(d\) on the day before the \({9}^{th}\) Christmas.
(d) Tanaka desired for the amount in her account to be at least €\(17\ 000\) the day before her brother turned 9 years old. Determine the minimum value of the monthly deposit required to achieve this. Give your answer correct to the nearest euro.
(e) As soon as the \({9}^{th}\) Christmas passed, Tanaka decided to invest €\(7\ 329\) of this money in an account of the same type earning 0.44% interest per annum. She withdraws €\(470\) every year on her Christmas day to buy her brother a present. Determine how many years and months it will take until there is no money in the account.
(a) Find an expression for \(D_{1}\) and show that \(D_{2} = {1.0032}^{2}d + 1.0032d\).
Recall that for an increase, the factor must be greater than 1, and for a decrease, the factor must be less than 1. Hence, the 0.32% interest is a factor which is greater than 1, as the amount in the account increases.
\[D_{1} = d(1 + 0.0032) = \boxed{{d}\left( {1.0032} \right)}\]
[1]
It is clear that the interest is compounded, meaning that the interest of the previous amount is calculated and added, using the previous amount as well.
\begin{align*}D_{2} &= D_{1} + D_{1}(1.0032)\\D_{2} &= d(1.0032) + d(1.0032)(1.0032)\\{D}_{{2}} &= \boxed{{d}\left( {1.0032} \right){+ d}\left( {1.0032} \right)^{{2}}}\\\end{align*}
[1]
(b) (i) Find a similar expression for \(D_{3}\) and \(D_{4}\).
\begin{align*}D_{3} &= D_{1} + D_{2} + D_{2}(1.0032)\\{D}_{{3}} &= \boxed{{d(1.0032) + d}\left( {1.0032} \right)^{{2}}{+ d}\left( {1.0032}\right)^{{3}}}\\\end{align*}
[1]
\begin{align*}D_{4} &= D_{1} + D_{2} + D_{3} + D_{4}(1.0032)\\{D}_{{4}} &= \boxed{{d}\left( {1.0032} \right){+ d}\left( {1.0032} \right)^{{2}}{+ d}\left( {1.0032} \right)^{{3}}{+ d}\left( {1.0032} \right)^{{4}}}\\\end{align*}
[1]
(b) (ii) Hence, show that the amount in Tanaka's account the day before her brother's \({7}^{th}\) birthday is given by \(313.5({1.0032}^{83} - 1)d\).
\({7}^{th}\) birthday means that (almost) 7 years have passed.
\[7\ \left( \text{years} \right) = 7 \times 12\ \left( \text{months} \right) = 84\ \left( \text{months} \right)\]
However, it must be kept in mind that the amount being calculated is of the day before Tanaka's brother's \({7}^{th}\) birthday. Therefore, the interest has not been calculated yet for the last month of the year, as December has not ended yet (\({24}^{th}\) December).
\[n = 84 - 1\ \left( \text{months} \right) = 83\ (\text{months})\]
[1]
Recognize the general pattern and factor:
\begin{align*}D_{n} &= d(1.0032) + d(1.0032)^{2} + \cdots + d(0.0032)^{n}\\D_{n} &= d\lbrack(1.0032) + (1.0032)^{2} + \cdots + (0.0032)^{n}\rbrack\\\end{align*}
Notice the geometric series summation inside the brackets \({[}{]}\):
\[D_{n} = d\lbrack S_{n}\rbrack\]
\[S_{n} = (1.0032) + (1.0032)^{2} + \cdots + (0.0032)^{n}\]
Where \(u_{1} = 1.0032\), and \(r = 1.0032\).
Therefore, using the geometric series summation formula in the formula booklet:
\begin{align*}S_{n} & = \frac{u_{1}\left( r^{n} - 1 \right)}{r - 1}\\S_{83} & = \frac{1.0032\left( {1.0032}^{83} - 1 \right)}{1.0032 - 1}\\S_{83} & = \frac{1.0032\left( {1.0032}^{83} - 1 \right)}{0.0032}\\S_{83} & = 313.5\left( {1.0032}^{83} - 1 \right)\\S_{83} & = 313.5\left( {1.0032}^{83} - 1 \right)\\\end{align*}
[1]
Hence:
\begin{align*}D_{n} &= d\left\lbrack S_{n} \right\rbrack\\D_{83} &= d\left\lbrack S_{83} \right\rbrack\\D_{83} &= d\left\lbrack 313.5\left( {1.0032}^{83} - 1 \right) \right\rbrack\\{D}_{{83}} &= \boxed{{313.5}\left( {1.0032}^{{83}}{- 1} \right){d}}\\\end{align*}
(c) Write down an expression for \(D_{n}\) in terms of \(d\) on the day before the \({9}^{th}\) Christmas.
\({9}^{th}\) Christmas is the same as the birthday, meaning that (almost) 9 years have passed.
\begin{align}9\ \left( \text{years} \right) &= 9 \times 12\ \left( \text{months} \right)\\9\ \left( \text{years} \right) &= 108\ \left( \text{months} \right)\\\end{align}
However, it must be kept in mind that the amount being calculated is of the day before Tanaka's brother's \({9}^{th}\) birthday. Therefore, the interest has not been calculated yet for the last month of the year, as December has not ended yet \({24}^{th}\) December).
\[n = 108 - 1\ \left( \text{months} \right) = 107\ (\text{months})\]
[1]
Hence:
\begin{align}D_{n} &= 313.5\left( {1.0032}^{n} - 1 \right)d\\D_{(9 \times 12) - 1} &= 313.5\left( {1.0032}^{(9 \times 12) - 1} - 1 \right)d\\{D}_{{107}} &= \boxed{{313.5}\left( {1.0032}^{{107}}{- 1} \right){d}}\\\end{align}
[1]
(d) Tanaka desired for the amount in her account to be at least €\(17\ 000\) the day before her brother turned 9 years old. Determine the minimum value of the monthly deposit required to achieve this. Give your answer correct to the nearest euro.
Recall that the number \({n}\) of months passed before Tanaka's brother turned 9 years old is given by \({n}\) = 107 (as calculated in part (c)).
Therefore:
\begin{align*}17000 &< {D}_{107}\\17000 &< 313.5\left( {1.0032}^{107} - 1 \right)d\\\end{align*}
[1]
Using G.D.C. solver or through algebraic manipulation to solve for \(d\):
\[d = \boxed{134}\]
[3] (accept \(d\) \(\geq\) €\(134\))
(e) As soon as the \({9}^{th}\) Christmas passed, Tanaka decided to invest €\(7\ 329\) of this money in an account of the same type earning 0.44% interest per annum. She withdraws €\(470\) every year on her Christmas day to buy her brother a present. Determine how many years and months it will take until there is no money in the account.
The money in the account after \({n}\) years without taking any money out is given by \(7329({r)}^{n}\).
Letting some variable \(E_{n}\) represent the money in the account, where some variable \(F_{n}\) is the total amount taken out:
\[E_{n} = 7329{(1.0044)}^{n} - F_{n}\]
[1]
As €470 is taken out every year, taking interest into account:
\begin{align*}F_{n} &= 470(1.0044) + 470(1.0044)^{2} + \cdots + 470(1.0044)^{n}\\F_{n} &= 470\lbrack(1.0044) + (1.0044)^{2} + \cdots + (1.0044)^{n}\rbrack\\\end{align*}
Notice the geometric series summation inside the brackets \({[}{]}\):
\[F_{n} = 470\lbrack S_{n}\rbrack\]
\[S_{n} = (1.0044) + (1.0044)^{2} + \cdots + (0.0044)^{n}\]
Where \(u_{1} = 1.0032\), and \(r = 1.0032\). Therefore, using the geometric series summation formula in the formula booklet:
\begin{align*}S_{n} &= \frac{u_{1}\left( r^{n} - 1 \right)}{r - 1}\\S_{n} &= \frac{1.0044\left( {1.0044}^{n} - 1 \right)}{1.0044 - 1}\\S_{n} &= \frac{1.0044\left( {1.0032}^{83} - 1 \right)}{0.0044}\\S_{n} &= 228.273\left( {1.0044}^{n} - 1 \right)\\\end{align*}
[1]
Hence:
\begin{align*}F_{n} &= 470\lbrack S_{n}\rbrack\\F_{n} &= 470\left\lbrack 228.273\left( {1.0044}^{n} - 1 \right) \right\rbrack\\F_{n} &= 228.273\left( {1.0044}^{n} - 1 \right)470\\\end{align*}
Therefore, the entire equation quantifying the total money in the new account after \({n}\) years:
\begin{align*}E_{n} &= 7329(1.0044)^{n} - F_{n}\\E_{n} &= 7329{(1.0044)}^{n} - 228.273\left( {1.0044}^{n} - 1 \right)470\\\end{align*}
Equate \(E_{n}\) to 0, as the number of years/months \({n}\) for when the account balance reaches 0 is being found:
\[E_{n} = 0 = 7329(1.0044)^{n} - 228.273\left( {1.0044}^{n} - 1 \right)470\]
[1]
Using G.D.C. solver or through algebraic manipulation to solve for \({n}\):
\[n = 16.12\ (\text{months})\]
[3]
To convert from months to years and months:
\begin{align*}n &= 16.12\ \left( \text{months} \right) = \frac{16.12}{12}\ \left( \text{years} \right) = 1.34\ \left( \text{years} \right)\\n &= 1\ \left( \text{year} \right)\ \text{and}\ 0.34\ \left( \text{years} \right)\\\end{align*}
[1]
\[{0.34\ \left( \text{years} \right) = 0.34 \times 12\ \left( \text{months} \right) = 4\ (\text{months})}\]
Hence:
\[n = \boxed{\text{1 year and 4 months}}\]
[1]
(a) Show that \({x}^{3}+2{x}^{2}-5x-10=({x}^{2}-5)(x+2)\)
(b) Find the coefficient of \({x}^{6}\) in the expansion of \({(x+2)}^{2}{({x}^{3}+2{x}^{2}-5x-10)}^{3}\)
(a) Show that \({x}^{3}+2{x}^{2}-5x-10=({x}^{2}-5)(x+2)\)
We can simply factor the LHS of the equation to give us the RHS, demonstrating that they are equivalent.
\begin{align*}
\text{LHS}&={x}^{3}+2{x}^{2}-5x-10\\
&={x}^{2}(x+2)-5(x+2)\\
&=\boxed{({x}^{2}-5)(x+2)=\text{RHS}}\\
\end{align*}
[1]
(b) Find the coefficient of \({x}^{3}\) in the expansion of \({(x+2)}^{2}{({x}^{3}+2{x}^{2}-5x-10)}^{3}\)
Using the equivalency in part (a), we can rewrite the expression as:
\begin{align*}
{(x+2)}^{2}{({x}^{3}+2{x}^{2}-5x-10)}^{3}&={(x+2)}^{2}{[({x}^{2}-5)(x+2)]}^{3}\\
&={(x+2)}^{2}{({x}^{2}-5)}^{3}{(x+2)}^{3}\\
&={(x+2)}^{(2+3)}{({x}^{2}-5)}^{3}\\
&={(x+2)}^{5}{({x}^{2}-5)}^{3}\\
\end{align*}
Then through inspection, it can be identified that there are only 3 combinations of products of \(x\) yielding \({x}^{6}\) with the given binomials:
\[{x}^{0}×{x}^{6}={x}^{6}\]
\[\text{OR}\]
\[{x}^{2}×{x}^{4}={x}^{6}\]
\[\text{OR}\]
\[{x}^{4}×{x}^{2}={x}^{6}\]
Hence, using combinatorics and the binomial theorem, these combinations can be added up to provide the coefficient on \({x}^{6}\):
\[[(^{5}{C}_{0}){(x)}^{0}{(2)}^{5})×((^{3}{C}_{3}){({x}^{2})}^{3}{(-5)}^{0})]\]
\[\text{OR}\]
\[[(^{5}{C}_{2}){(x)}^{2}{(2)}^{3})×((^{3}{C}_{2}){({x}^{2})}^{2}{(-5)}^{1})]\]
\[\text{OR}\]
\[[(^{5}{C}_{4}){(x)}^{4}{(2)}^{1})×((^{3}{C}_{1}){({x}^{2})}^{1}{(-5)}^{2})]\]
Adding these values up and simplifying using a GDC yields:
\[[(^{5}{C}_{0}){(x)}^{0}{(2)}^{5})×((^{3}{C}_{3}){({x}^{2})}^{3}{(-5)}^{0})]+[(^{5}{C}_{2}){(x)}^{2}{(2)}^{3})×((^{3}{C}_{2}){({x}^{2})}^{2}{(-5)}^{1})]+[(^{5}{C}_{4}){(x)}^{4}{(2)}^{1})×((^{3}{C}_{1}){({x}^{2})}^{1}{(-5)}^{2})]=-418{x}^{6}\]
Hence:
\[\boxed{\text{Coefficient: }-418}\]
[6]
[Worksheets] for everyday use
