1

Use the binomial theorem to find the coefficient for $x^3$ on both sides of the expansion of:

$(1+x)^3$$(1+x)^3$ $=$ $(1+x)^6$

i. Hence show $(_3C_0)^2+(_3C_1)^2+(_3C_2)^2+(_3C_3)^2$ $=$ $_6C_3$

ii. Use the same argument with $(1+x)^n(1+x)^n = (1+x)^{2n}$ to prove

$\sum_{k=0}^n(_nC_k)^2=_{2n}C_n$

Thank you.

Martin Sleziak
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Tom Walker
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1 Answers1

2

It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. We also use the notation $\binom{n}{k}$ instead of $_nC_k$.

We obtain \begin{align*} [x^3](1+x)^3(1+x)^3&=[x^3]\left(\sum_{j=0}^3\binom{3}{j}x^j\right)\left(\sum_{k=0}^3\binom{3}{k}x^k\right)\tag{1}\\ &=\sum_{j=0}^3\binom{3}{j}[x^{3-j}]\sum_{k=0}^3\binom{3}{k}x^k\tag{2}\\ &=\sum_{j=0}^3\binom{3}{j}\binom{3}{3-j}\tag{3}\\ &=\sum_{j=0}^3\binom{3}{j}^2\tag{4}\\ \end{align*}

on the other hand we obtain \begin{align*} [x^3](1+x)^3(1+x)^3&=[x^3](1+x)^6\\ &=[x^3]\sum_{j=0}^6\binom{6}{j}x^j\tag{5}\\ &=\binom{6}{3} \end{align*}

Comment:

  • In (1) we apply the binomial theorem twice.

  • In (2) we use the linearity of the coefficient of operator and apply the rule \begin{align*} [x^p]x^qA(x)=[x^{p-q}]A(x) \end{align*} to the summands of the left series.

  • In (3) we select the coefficient of $x^{3-j}$ from the right series.

  • In (4) we use the symmetry $\binom{n}{k}=\binom{n}{n-k}$.

  • In (5) we again apply the binomial theorem and select the coefficient of $x^3$.

We conclude \begin{align*} \sum_{j=0}^3\binom{3}{j}^2=\binom{6}{3}\\ \end{align*}

The calculation to show \begin{align*} \sum_{j=0}^n\binom{n}{j}^2=\binom{2n}{n}\\ \end{align*} can be done analogously.

Markus Scheuer
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