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A polynomial in $x$ is defined by $$a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}=(x+2x^2+ \cdots +nx^n)^2.$$ Show that the sum of all $a_i$, for $i\in\{n+1,n+2, \ldots , 2n\}$, is $$ \frac {n(n+1)(5n^2+5n+2)} {24}.$$

I don't know how to proceed. I know the Multinomial theorem, however, I have problems in applying it. Any help will be appreciated as it will help me understand the theorem well.

Thanks!

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  • $\begingroup$ I have the exact same question. $\endgroup$
    – Boka Peer
    Dec 22 '19 at 5:09
  • $\begingroup$ @MMM Sorry, missed a 2. Thanks for pointing out. $\endgroup$
    – Sen47
    Dec 22 '19 at 5:11
  • $\begingroup$ Do you need to use the multinomial theorem for this question? I can't spot a neat way to apply the multinomial theorem right now but let me know if you would be happy for an answer using generating functions. $\endgroup$ Dec 22 '19 at 6:46
  • $\begingroup$ @RagibZaman Sure sir! $\endgroup$
    – Sen47
    Dec 22 '19 at 8:36
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Here is an easy method using multinomial coefficients. Put $x=1$ to get the sum of all the coefficients. Now, we want to evaluate $\sum_{i=0}^na_i$, then we will subtract that from sum of all coefficients. Observe that these coefficients will remain unaltered even in the following expansion (because the additional terms do not contribute to powers less than $x^{n+1}$): $$(x+2x^2+3x^3+...)^2 = x^2(1+2x+3x^2+...)^2$$ $$ = x^2\Bigg(\frac{1}{(1-x)^2}\Bigg)^2$$ $$ = \frac{x^2}{(1-x)^4}$$ $$ = x^2\sum_{m=0}^\infty\binom{m+4-1}{4-1}x^m$$ Now, apply the identity that $$\sum_{i=k}^n\binom{i}{k} = \binom{n+1}{k+1}$$ and you are done.

Hope it helps:)

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  • $\begingroup$ Where did you use the multinomial theorem? $\endgroup$ Dec 22 '19 at 12:59
  • $\begingroup$ The coefficient of $x^m$ in expansion of $\frac{1}{(1-x)^4}$ is multinomial coefficient. $\endgroup$
    – Martund
    Dec 22 '19 at 13:20
  • $\begingroup$ It seems like you are calling the identity $\frac{1}{(1-x)^{n+1}} = \sum_{m=0}^{\infty} \binom{m+n}{n} x^m$ the multinomial theorem, which actually refers to the identity $(x_1 + x_2 + \ldots + x_m)^n = \sum_{k_1 + k_2 + \ldots + k_m = n} \binom{n}{k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m}.$ $\endgroup$ Dec 22 '19 at 14:31
  • $\begingroup$ I know that multinomial theorem is that which you are saying. The coefficients I am using is the number of multinomial coefficients used in multinomial expansion. $\endgroup$
    – Martund
    Dec 22 '19 at 14:37
  • $\begingroup$ @RagibZaman, I have edited the answer, if use of the term enraged you :P $\endgroup$
    – Martund
    Dec 22 '19 at 14:39
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Write $(x+2x^2+\cdots+nx^n)^2=(x+2x^2+\cdots+nx^n)(x+2x^2+\cdots+nx^n)$ and for each coefficient in the first factor find the sum of the coefficients in the second factor with which it will enter into the desired sum: The coefficient $k$ in the first factor is paired with $n+1-k$ through $n$ in the second factor, so it contributes

\begin{eqnarray*} k\sum_{j=n+1-k}^nj &=& k\left(\sum_{j=1}^nj-\sum_{j=1}^{n-k}j\right) \\ &=& k\left(\frac{n(n+1)}2-\frac{(n-k)(n-k+1)}2\right) \\&=& \left(n+\frac12\right)k^2-\frac12k^3\;. \end{eqnarray*}

Then summing over $k$ yields

\begin{eqnarray*} \sum_{k=1}^n\left(\left(n+\frac12\right)k^2-\frac12k^3\right) &=&\left(n+\frac12\right)\sum_{k=1}^nk^2-\frac12\sum_{k=1}^nk^3 \\ &=& \left(n+\frac12\right)\frac{n(n+1)(2n+1)}6-\frac12\left(\frac{n(n+1)}2\right)^2 \\ &=& \frac{n(n+1)(5n^2+5n+2)}{24}\;. \end{eqnarray*}

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Let $S_n$ be the required sum. By expanding the right hand side of $$a_0 + a_1 x + \ldots + a_{2n} x^{2n} = (x + 2x^2 + \ldots + nx^n)^2$$ we have

$$a_{n+i} = n \cdot i + (n-1) \cdot (i+1) + (n-2) \cdot (i+2) + \cdots \ + i \cdot n$$ for $i=1, 2, \ldots, n.$

Summing over $i=1,2,\ldots, n$ gives

$$ S_n = n(1 + 2 + 3+ \ldots + n) + (n-1)(2+ 3 + \ldots + n) + \ \ldots \ + 2( (n-1) + n) + 1(n) $$

The terms in the brackets are the sum of the first $n$ integers (which equals $\binom{n+1}{2}$) minus the sum of the first $k$ integers (which equals $\binom{k+1}{2}$) so we get

$$ S_n = \sum_{k=0}^{n-1} (n-k) \left( \binom{n+1}{2} - \binom{k+1}{2} \right)$$

$$ = \binom{n+1}{2} \sum_{k=0}^{n-1} (n-k) - \sum_{k=1}^{n-1} \binom{n-k}{1} \binom{k+1}{2}$$

Again, we have $\sum_{k=0}^{n-1} (n-k) = \binom{n+1}{2}$ so the first term above simplifies to $\binom{n+1}{2}^2.$

To pick a $4$-element subset from $n+2$ elements we follow this scheme - Pick element $k+2$ ($ \ k$ from $1$ to $n-1$) to be $3$-rd element of the subset, then pick $2$ from the $k+1$ elements on its left and $1$ element from the $n-k$ on its right. Summing over all possibilities of what the $3$rd element could be yields the second term above.

Therefore, $$S_n = \binom{n+1}{2}^2 - \binom{n+2}{4}$$ $$ \ \ \ \ \ \ \ \ \ = \frac {n(n+1)(5n^2+5n+2)} {24}$$

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You can check in the link it's similar to what your question is. https://www.mathsdiscussion.com/forum/topic/sum-of-coefficients/?part=1#postid-55

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