Question

Coefficient of $x^{18}$ in
${(1+x+2x^2+3x^3+.....+18x^{18})}^2$ is equal to

Answer 1

Lets find coefficient of $x^{18}$ in,

${(1+x+2x^2+3x^3+.....+18x^{18})}^2$

$=(1+x+2x^2+...+18x^{18})\times (1+x+2x^2+....+18x^{18})$

To find coefficient of $x^{18}$,

take the product of $rth$ term in the first sequence and the

$(n-r+1)th$ term in the second sequence.

In first step, product by $1$ need to finish.

i.e., $1\times 18x^{18}$ and $18x^{18}\times 1$

it gives the coefficient is $18+18=36$

For the remaining sum of the terms apply the concept of

${\sum }_{r=1}^{n}r(n-r+1)$

In each expansion, $19$ terms are available. in which $2$ terms are evaluated.

therefore, the remaining terms are $17$.

$\Rightarrow n=17$

Now, lets take $(1+x+2x^2+...+18x^{18})\times (1+x+2x^2+....+18x^{18})$

$=36+{\sum }_{r=1}^{17}r(17-r+1)$

$=36+{\sum }_{r=1}^{17}r(18-r)$

$=36+{\sum }_{r=1}^{17}(18r-r^2)$

$=36+18{\sum }_{r=1}^{17}r-{\sum }_{r=1}^{17}{r}^2$

$=36+18\left(\frac{17\times (17+1)}{2}\right)-\left(\frac{17\times (17+1)\times 2\left(17\right)+1}{6}\right)$

[ $\because \sum n=\frac{n\times (n+1)}{2},\sum n^2=\frac{n\times (n+1)\times (2n+1)}{6}$]

$=36+(9\times 17\times 18)-(17\times 3\times 35)$

$=36+2754-1785$

$=1005$

Therefore, the coefficient of $x^{18}$ in ${(1+x+2x^2+3x^3+.....+18x^{18})}^2$ is $1005$.