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Question: The coefficient of \({{X}^{n}}\) in \({{(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})}^{2}}\). A...

The coefficient of Xn{{X}^{n}} in (1+x+2x2+3x3+........+nxn)2{{(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})}^{2}}.
A. n(n2+11)6\dfrac{n\left( {{n}^{2}}+11 \right)}{6}
B. n(n2+10)6\dfrac{n\left( {{n}^{2}}+10 \right)}{6}
C. n(n2+11)4\dfrac{n\left( {{n}^{2}}+11 \right)}{4}
D. n(n2+10)4\dfrac{n\left( {{n}^{2}}+10 \right)}{4}

Explanation

Solution

To solve the given question use Binomial theorem. According to the Binomial theorem, for any positive integer nn, the nth{{n}^{th}} power of the sum of the two numbers. X and y may be expressed as the sum of n+1n+1 terms of the forms
(x+y)n=∑r=0nCrnxn−ryr{{(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 xn{{x}^{n}} in (1+x+2x2+3x3+........+nxn)2{{(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})}^{2}}.
We can write the above expression as:
(1+x+2x2+3x3+........+nxn)′(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})' (1+x+2x2+3x3+........+nxn)′(1+x+2{{x}^{2}}+3{{x}^{3}}+........+n{{x}^{n}})'
Expanding the above bracket, we get:
(1+x+2x2+3x3+........+(n−1)xn−1+nxn)′(1+x+2{{x}^{2}}+3{{x}^{3}}+........+(n-1){{x}^{n-1}}+n{{x}^{n}})' (1+x+2x2+3x3+........+(n−1)xn−1+nxn)′(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 xn{{x}^{n}} is equal to :-
n+(n−1)+(n−1)+2(n−2)+3(n−3)+..........+(n−1).1+nn+(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 :-
∑r=1nr(n−r)+2n\sum\limits_{r=1}^{n}{r(n-r)}+2n
Solve the above equation by multiplying r with the bracket.
⇒∑r=1nr(n−r)+2n=∑r=1n(rn−r2)+2n\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
∑r=1nr(n−r)+2n=∑r=1nrn−∑r=1nr2+2n..........(A)\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.
∑r=1nr=(n)(n+1)2\sum\limits_{r=1}^{n}{r}=\dfrac{(n)(n+1)}{2}
⇒∑r=1n(r)2=n(n+1)(2n+1)6\Rightarrow\sum\limits_{r=1}^{n}{{{(r)}^{2}}}=\dfrac{n(n+1)(2n+1)}{6}
Substituting the values of ∑r=1nnr\sum\limits_{r=1}^{n}{nr} and ∑r=1nr2\sum\limits_{r=1}^{n}{{{r}^{2}}} in (A)\left( A \right)
So the equation becomes: n(n)(n+1)2\dfrac{n(n)(n+1)}{2} -n(n+1)(2n+1)6+2n\dfrac{n(n+1)(2n+1)}{6}+2n
Taking common from the above term we get
n(n+1)2[n−2n+13]+2n\dfrac{n(n+1)}{2}\left[ n-\dfrac{2n+1}{3} \right]+2n ………….(B)
Taking L.C.M. of the term:
[n−2n+13]\left[ n-\dfrac{2n+1}{3} \right] we get 3n−2n−13\dfrac{3n-2n-1}{3} = n−13\dfrac{n-1}{3}
Substituting the above value in equation B
n+12[n−13]+2n\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)
n2−1=(n−1)(n+1){{n}^{2}}-1=(n-1)(n+1)

And also multiply the denominator
2×3=62\times 3=6
⇒n(n2−1)6+2n\Rightarrow\dfrac{n({{n}^{2}}-1)}{6}+2n
By solving the term n(n2−1)=n3−nn({{n}^{2}}-1)={{n}^{3}}-n
So, the above equation becomes.
n3−n6+2n\dfrac{n{}^{3}-n}{6}+2n
Taking L.C.M. of the above term and we get
n3−n+12n6\dfrac{n{}^{3}-n+12n}{6}
As n + 12nn\text{ }+\text{ }12n are having the same power. So get subtracted we get
Substituting in above = (n3+11n)6\dfrac{(n{}^{3}+11n)}{6}
Again, taking n common from the above equation.
n(n2+11)6\dfrac{n(n{}^{2}+11)}{6}
Hence proved the result.

Note: For solving the expansion of series in binomial. Using binomial series term as nCrxn−1yr{}^{n}C{}_{r}{{x}^{n-1}}{{y}^{r}} and also nCr=n!r!(n−r)!{}^{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.