Binomial help (1 Viewer)

petergriffin198 · Oct 9, 2015

Hsc 2013 mathematics extention 1 question 14b (iii)

These r the solutions(partial) . I dont get it from here some1 explain / diffferent method plz

InteGrand · Oct 9, 2015

$Note that $\binom{4n}{2n}$ is the coefficient $C_{2n}$ of $x^{2n}$ in the expansion of:$

$(1+x)^{4n}=\left((1+x)^2\right)^{2n}$

$=(1+2x+x^2)^{2n} = (1+x(x+2))^{2n}$

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

$$=\sum_{k=0}^{2n}\left(\binom{2n}{k}\sum_{j=0}^{2n-k} \binom{2n-k}{j}2^{2n-k-j}x^{2n-k+j} \right)$ (using the given fact in (iii))$

$Now, the coefficient of $x^{2n}$ in this last expression must also be $C_{2n}=\binom{4n}{2n}$. But the coefficient of $x^{2n}$ in this last expression is found by simply taking the one and only $j=k$ term for each $k=0,1,2,...,n$, since then the $x$ term has exponent $2n-k+k$ (as $k=j$) $= 2n$, as desired. This coefficient in the second sum is $\binom{2n-k}{k}2^{2n-2k}$ (due to replacing $j$'s with $k$ as we are only taking the $k=j$ term).Note that since for each $k=0,1,...,2n$, we have $j\in \{0,1,...,2n-k\}$, so once $k$ is $n+1$ or bigger, we can no longer achieve $j=k$, as $j$ only goes up to $2n-k$, which would be smaller than $n$ if $k\geq n+1$. This implies that the coefficient of $x^{2n}$ in the last expression is:$

$$=\sum_{k=0}^{n}\binom{2n}{k}\binom{2n-k}{k}2^{2n-2k}$ i.e. sum up coefficients of all $x^{2n}$ terms in the expansion to get the overall coefficient. Upon equating coefficients $C_{2n}$ we have$

$\binom{4n}{2n}=\sum_{k=0}^{n}\binom{2n}{k}\binom{2n-k}{k}2^{2n-2k}$. $\blacksquare$

Binomial help (1 Viewer)

petergriffin198

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InteGrand

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