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Question: The co-efficient of \({x^n}\) in \({\left( {1 + x + 2{x^2} + 3{x^3} + .... + n{x^n}} \right)^2}\) is...

The co-efficient of xn{x^n} in (1+x+2x2+3x3+....+nxn)2{\left( {1 + x + 2{x^2} + 3{x^3} + .... + n{x^n}} \right)^2} is:
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

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

Complete step-by-step answer:
We are given (1+x+2x2+3x3+....+nxn)2{\left( {1 + x + 2{x^2} + 3{x^3} + .... + n{x^n}} \right)^2}is equal to (1+x+2x2+3x3+....+nxn)(1+x+2x2+3x3+....+nxn)\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 xn{x^n}.
When we will multiply the power 1 with the power nn, power 2 with n−1n - 1 and so on, then we will get the coefficient of xn{x^n}, which is
The coefficient of xn{x^n} will be n+1(n−1)+2(n−2)+.........(n−1)(1)+nn + 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(n−1)+2(n−2)+.........(n−1)(1)+nn + 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, ∑r=1nr(n−r)+(2n)\sum\limits_{r = 1}^n {r\left( {n - r} \right) + \left( {2n} \right)}
Find the summation, using the formula for the sum of nn terms and n2{n^2}terms.
∑r=1nr(n−r)+(2n)=∑r=1nnr−r2+(2n)\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 nn natural numbers is given by n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2} and sum of square of first nn natural numbers is n(n+1)(2n+1)6\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}
Therefore, we have, ∑r=1nnr−r2+(2n)=(n)n(n+1)2−n(n+1)(2n+1)6+2n\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,
(n)n(n+1)2−n(n+1)(2n+1)6+(2n) =n(n+1)2[n−(2n+13)]+2n =n(n+1)2[3n−2n−13]+2n =n(n+1)2[n−13]+2n =n(n2−1)6+2n =n3−n+12n6 =n3+11n6 =n(n2+11)6  \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 xn{x^n} can be calculated using the am(an)=am+n{a^m}\left( {{a^n}} \right) = {a^{m + n}}. Sum of first nn natural numbers is given by n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2} and sum of square of first nn natural numbers is n(n+1)(2n+1)6\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}