Question

Coefficient of ${{x}^{18}}$ in ${{\left( 1+x+2{{x}^{2}}+3{{x}^{3}}+....+18{{x}^{18}} \right)}^{2}}$ is equal to?
(a) 995
(b) 1005
(c) 1235
(d) None of these

Answer 1

Hint: Write the expression ${{\left( 1+x+2{{x}^{2}}+3{{x}^{3}}+....+18{{x}^{18}} \right)}^{2}}$ as the product of two similar expressions by breaking the exponent of each expression into 1. Now, check which term from the first expression needs to be multiplied with the particular term of the second expression to get the exponent of x equal to 18. Form a summation series and use the formulas $\sum\limits_{1}^{n}{r}=\dfrac{r\left( r+1 \right)}{2}$ and $\sum\limits_{1}^{n}{{{r}^{2}}}=\dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}$ to get the answer.

Complete step by step answer:
Here we have been provided with the expression ${{\left( 1+x+2{{x}^{2}}+3{{x}^{3}}+....+18{{x}^{18}} \right)}^{2}}$ and we are asked to find the coefficient of ${{x}^{18}}$. We can write the above expression as the product of two similar terms with exponent of each as 1. So we get,
$\Rightarrow {{\left( 1+x+2{{x}^{2}}+....+18{{x}^{18}} \right)}^{2}}=\left( 1+x+2{{x}^{2}}+....+18{{x}^{18}} \right)\left( 1+x+2{{x}^{2}}+....+18{{x}^{18}} \right)$
Now, we can see that we will get ${{x}^{18}}$ when we will multiply the first term of expression 1 with the last term of expression 2, second term of first expression with second last term of second expression and similarly moving ahead till we reach to the product of last term of first expression with the first term of the second expression. While doing this we will get the sum of coefficients as:
$\Rightarrow \text{Sum}=\left( 1\times 18 \right)+\left( 1\times 17 \right)+\left( 2\times 16 \right)+\left( 3\times 15 \right)+...+\left( 8\times 10 \right)+\left( 9\times 9 \right)+\left( 10\times 8 \right)+...+\left( 18\times 1 \right)$
On simplifying we get,
\[\Rightarrow \text{Sum}=36+81+2\left[ \left( 1\times 17 \right)+\left( 2\times 16 \right)+\left( 3\times 15 \right)+...+\left( 8\times 10 \right) \right]\]
The terms inside the bracket can be written in summation form as:
\[\begin{align}
  & \Rightarrow \text{Sum}=117+2\left[ \sum\limits_{1}^{8}{r\left( 18-r \right)} \right] \\
 & \Rightarrow \text{Sum}=117+2\left[ \sum\limits_{1}^{8}{18r}-\sum\limits_{1}^{8}{{{r}^{2}}} \right] \\
 & \Rightarrow \text{Sum}=117+2\left[ 18\sum\limits_{1}^{8}{r}-\sum\limits_{1}^{8}{{{r}^{2}}} \right] \\
\end{align}\]
Using the formulas $\sum\limits_{1}^{n}{r}=\dfrac{r\left( r+1 \right)}{2}$ and $\sum\limits_{1}^{n}{{{r}^{2}}}=\dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}$ we get,
\[\begin{align}
  & \Rightarrow \text{Sum}=117+2\left[ 18\left( \dfrac{8\times 9}{2} \right)-\left( \dfrac{8\times 9\times 17}{2} \right) \right] \\
 & \therefore \text{Sum}=1005 \\
\end{align}\]

So, the correct answer is “Option b”.

Note: Note you cannot keep on multiplying all the terms or finding the square of the expression to get the answer because there are 19 terms in the expression here so it is not possible to do that. This is the reason we need to form a series like we formed above to simplify. Note that the formulas $\sum\limits_{1}^{n}{r}=\dfrac{r\left( r+1 \right)}{2}$ and $\sum\limits_{1}^{n}{{{r}^{2}}}=\dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}$ are the sum of first n natural numbers and squares of first n natural numbers respectively.