Paper 4, Section II, 7E7 \mathrm{E}

Numbers and Sets | Part IA, 2011

Define the binomial coefficient (ni)\left(\begin{array}{l}n \\ i\end{array}\right), where nn is a positive integer and ii is an integer with 0in0 \leqslant i \leqslant n. Arguing from your definition, show that i=0n(ni)=2n\sum_{i=0}^{n}\left(\begin{array}{l}n \\ i\end{array}\right)=2^{n}.

Prove the binomial theorem, that (1+x)n=i=0n(ni)xi(1+x)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\ i\end{array}\right) x^{i} for any real number xx.

By differentiating this expression, or otherwise, evaluate i=0ni(ni)\sum_{i=0}^{n} i\left(\begin{array}{l}n \\ i\end{array}\right) and i=0ni2(ni)\sum_{i=0}^{n} i^{2}\left(\begin{array}{l}n \\ i\end{array}\right). By considering the identity (1+x)n(1+x)n=(1+x)2n(1+x)^{n}(1+x)^{n}=(1+x)^{2 n}, or otherwise, show that

i=0n(ni)2=(2nn)\sum_{i=0}^{n}\left(\begin{array}{l} n \\ i \end{array}\right)^{2}=\left(\begin{array}{c} 2 n \\ n \end{array}\right)

Show that i=0ni(ni)2=n2(2nn)\sum_{i=0}^{n} i\left(\begin{array}{c}n \\ i\end{array}\right)^{2}=\frac{n}{2}\left(\begin{array}{c}2 n \\ n\end{array}\right)

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