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When trying to apply the binomial theorem to a problem, I found a solution that involves this line:

$(x+1)(x^2-2)^9 = \sum_{k=0}^9 { 9 \choose k } (-1)^{k-1}2^{9-k}(x^{2k}+2^{2k + 1})$

I know the binomial theorem and the formula:

$\sum_{k=0}^n {n \choose k } a^n b^{n-k}$

But I really can't see how it's applied.

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3 Answers 3

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Apply the binomial theorem to $(x^2 - 2)^9$ and rearrange: $$(x + 1) (x^2 - 2)^9 = \\ (x + 1) \cdot \sum_{k=0}^9 {9 \choose k} (-2)^{9 - k} (x^2)^k = \\ (x + 1) \cdot \sum_{k=0}^9 {9 \choose k} (-2)^{9 - k} x^{2 k} = \\ \sum_{k=0}^9 {9 \choose k} (-2)^{9 - k} (x + 1) x^{2 k} = \\ \sum_{k=0}^9 {9 \choose k} (-2)^{9 - k} (x^{2 k + 1} + x^{2 k}) = \\ \sum_{k=0}^9 {9 \choose k} (-1)^{9-k} 2^{9 - k} (x^{2 k + 1} + x^{2 k}) = \\ \sum_{k=0}^9 {9 \choose k} (-1)^{k-1} 2^{9 - k} (x^{2 k + 1} + x^{2 k}, $$ where in the last step we took into account $(-1)^{9-k} = (-1)^{k-1}$ because $9 - k$ and $k - 1$ have the same parity.

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We have $$\begin{align*} (x+1)(x^2-2)^9 &= (x+1) \sum_{k=0}^9 \binom{9}{k} (x^2)^k (-2)^{9-k} \\ &= \sum_{k=0}^9 \binom{9}{k} (x+1)x^{2k} (-1)^{9-k} 2^{9-k} \\ &= \sum_{k=0}^9 \binom{9}{k} (x^{2k+1} + x^{2k}) (-1)^9(-1)^{-k} 2^{9-k} \\ &= \sum_{k=0}^9 \binom{9}{k} (-(-1)^k) 2^{9-k} (x^{2k+1} + x^{2k}), \end{align*} $$ which is equivalent to the given identity.

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$$(x^2-2)^9 = \sum_{k=0}^9 { 9 \choose k } \left(x^2\right)^k(-2)^{9-k}$$ would be a fairly standard application of the binomial expansion.

This could be rewritten as $$(x^2-2)^9 = \sum_{k=0}^9 { 9 \choose k } (-1)^{k-1}2^{9-k}(x^{2k})$$ if you really wanted to. Getting from $(-1)^{9-k}$ to $(-1)^{k-1}$ depends on $9$ being odd.

Now multiply both sides by $x+1$, noting that $(x+1)x^{2k} = (x^{2k}+2^{2k + 1})$ to give the original expression $$(x+1)(x^2-2)^9 = \sum_{k=0}^9 { 9 \choose k } (-1)^{k-1}2^{9-k}(x^{2k}+2^{2k + 1}).$$

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