Question

The co-efficient of ${x^n}$ in ${\left( {1 + x + 2{x^2} + 3{x^3} + .... + n{x^n}} \right)^2}$ is:
A.$\dfrac{{n\left( {{n^2} + 11} \right)}}{6}$
B.$\dfrac{{n\left( {{n^2} + 10} \right)}}{6}$
C.$\dfrac{{n\left( {{n^2} + 11} \right)}}{4}$
D.$\dfrac{{n\left( {{n^2} + 10} \right)}}{4}$

Answer 1

Hint: When we will multiply the power 1 with the power $n$, power 2 with $n - 1$ and so on, then we will get the coefficient of ${x^n}$. Write the general term of the coefficient of $x$. Simplify the general term by using the formula, sum of first $n$natural numbers is given by $\dfrac{{n\left( {n + 1} \right)}}{2}$ and sum of square of first $n$ natural numbers is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$

Complete step-by-step answer:
We are given ${\left( {1 + x + 2{x^2} + 3{x^3} + .... + n{x^n}} \right)^2}$is equal to $\left( {1 + x + 2{x^2} + 3{x^3} + .... + n{x^n}} \right)\left( {1 + x + 2{x^2} + 3{x^3} + .... + n{x^n}} \right)$
We have to find the coefficient of ${x^n}$.
When we will multiply the power 1 with the power $n$, power 2 with $n - 1$ and so on, then we will get the coefficient of ${x^n}$, which is
The coefficient of ${x^n}$ will be $n + 1\left( {n - 1} \right) + 2\left( {n - 2} \right) + .........\left( {n - 1} \right)\left( 1 \right) + n$
Now, we will simplify the expression, $n + 1\left( {n - 1} \right) + 2\left( {n - 2} \right) + .........\left( {n - 1} \right)\left( 1 \right) + n$ .
The general term for the expression can be written as, \[\sum\limits_{r = 1}^n {r\left( {n - r} \right) + \left( {2n} \right)} \]
Find the summation, using the formula for the sum of $n$ terms and ${n^2}$terms.
\[\sum\limits_{r = 1}^n {r\left( {n - r} \right) + \left( {2n} \right)} = \sum\limits_{r = 1}^n {nr - {r^2} + \left( {2n} \right)} \]
Sum of first $n$ natural numbers is given by $\dfrac{{n\left( {n + 1} \right)}}{2}$ and sum of square of first $n$ natural numbers is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
Therefore, we have, \[\sum\limits_{r = 1}^n {nr - {r^2} + \left( {2n} \right)} = \left( n \right)\dfrac{{n\left( {n + 1} \right)}}{2} - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 2n\]
On simplifying we get,
$
  \left( n \right)\dfrac{{n\left( {n + 1} \right)}}{2} - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \left( {2n} \right) \\
 = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {n - \left( {\dfrac{{2n + 1}}{3}} \right)} \right] + 2n \\
 = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{3n - 2n - 1}}{3}} \right] + 2n \\
 = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{n - 1}}{3}} \right] + 2n \\
 = \dfrac{{n\left( {{n^2} - 1} \right)}}{6} + 2n \\
 = \dfrac{{{n^3} - n + 12n}}{6} \\
 =\dfrac{{{n^3} + 11n}}{6} \\
  =\dfrac{{n\left( {{n^2} + 11} \right)}}{6} \\
$
Therefore, option A is correct.

Note: The coefficient of ${x^n}$ can be calculated using the ${a^m}\left( {{a^n}} \right) = {a^{m + n}}$. Sum of first $n$ natural numbers is given by $\dfrac{{n\left( {n + 1} \right)}}{2}$ and sum of square of first $n$ natural numbers is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$