Question

The coefficient of ${{X}^{n}}$ in ${{(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})}^{2}}$.
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: To solve the given question use Binomial theorem. According to the Binomial theorem, for any positive integer $n$, the ${{n}^{th}}$ power of the sum of the two numbers. X and y may be expressed as the sum of $n+1$ terms of the forms
${{(x+y)}^{n}}=\sum\limits_{r=0}^{n}{C_{r}^{n}{{x}^{n-r}}{{y}^{r}}}$

Complete step by step answer:
We have to coefficient of ${{x}^{n}}$ in ${{(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})}^{2}}$.
We can write the above expression as:
$(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})'$$(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})'$
Expanding the above bracket, we get:
$(1+x+2{{x}^{2}}+3{{x}^{3}}+........+(n-1){{x}^{n-1}}+n{{x}^{n}})'$$(1+x+2{{x}^{2}}+3{{x}^{3}}+........+(n-1){{x}^{n-1}}+n{{x}^{n}})'$
On multiplying the brackets we get that the coefficient of ${{x}^{n}}$ is equal to :-
$n+(n-1)+(n-1)+2(n-2)+3(n-3)+..........+(n-1).1+n$ so on………
Now, if we general the above expression in the form of a summation we get it as :-
$\sum\limits_{r=1}^{n}{r(n-r)}+2n$
Solve the above equation by multiplying r with the bracket.
$\Rightarrow \sum\limits_{r=1}^{n}{r(n-r)}+2n=\sum\limits_{r=1}^{n}{(rn-{{r}^{2}})}+2n$
Now, we can use the properties of summation and we get that
$\sum\limits_{r=1}^{n}{r(n-r)}+2n=\sum\limits_{r=1}^{n}{rn-}\sum\limits_{r=1}^{n}{{{r}^{2}}}+2n..........\left( A \right)$

Now, we can use the following identities to simplify the above equation.
$\sum\limits_{r=1}^{n}{r}=\dfrac{(n)(n+1)}{2}$
$\Rightarrow\sum\limits_{r=1}^{n}{{{(r)}^{2}}}=\dfrac{n(n+1)(2n+1)}{6}$
Substituting the values of $\sum\limits_{r=1}^{n}{nr}$ and $\sum\limits_{r=1}^{n}{{{r}^{2}}}$ in $\left( A \right)$
So the equation becomes: $\dfrac{n(n)(n+1)}{2}$ -$\dfrac{n(n+1)(2n+1)}{6}+2n$
Taking common from the above term we get
$\dfrac{n(n+1)}{2}\left[ n-\dfrac{2n+1}{3} \right]+2n$ ………….(B)
Taking L.C.M. of the term:
$\left[ n-\dfrac{2n+1}{3} \right]$ we get $\dfrac{3n-2n-1}{3}$ = $\dfrac{n-1}{3}$
Substituting the above value in equation B
$\dfrac{n+1}{2}\left[ \dfrac{n-1}{3} \right]+2n$ ……………..(C)
Expanding the terms by multiply and using a2 – b2 = (a – b)(a + b)
${{n}^{2}}-1=(n-1)(n+1)$

And also multiply the denominator
$2\times 3=6$
$\Rightarrow\dfrac{n({{n}^{2}}-1)}{6}+2n$
By solving the term $n({{n}^{2}}-1)={{n}^{3}}-n$
So, the above equation becomes.
$\dfrac{n{}^{3}-n}{6}+2n$
Taking L.C.M. of the above term and we get
$\dfrac{n{}^{3}-n+12n}{6}$
As $n\text{ }+\text{ }12n$ are having the same power. So get subtracted we get
Substituting in above = $\dfrac{(n{}^{3}+11n)}{6}$
Again, taking n common from the above equation.
$\dfrac{n(n{}^{2}+11)}{6}$
Hence proved the result.

Note: For solving the expansion of series in binomial. Using binomial series term as ${}^{n}C{}_{r}{{x}^{n-1}}{{y}^{r}}$ and also ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ as the general term. In the given question, having a coefficient of xn then using general terms to solve it.