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let $$f(x)=\sum_{i=0}^{k}(2x)^i(x+1)^{k+n-i}$$

show that $$x^{k+1}|[(x-1)f(x)+(x+1)^{k+n+1}]$$

my idea:since $$f(x)=-\dfrac{(x+1)^n(-2^{k+1}x^{k+1}+x(x+1)^k+(x+1)^k)}{x-1}$$ so $$(x-1)f(x)+(x+1)^{k+n+1}=(x+1)^{k+n+1}-(x+1)^{n+k}-x(x+1)^{n+k}-2^{k+1}x^{k+1}(x+1)^n$$

math110
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  • Looks good, however, why the double negation, and why did you a partial expansion of $(x+1)^{k+1}$ in the result of $\sum_{i=0}^k a^ib^{k-i}=\frac{a^{k+1}-b^{k+1}}{a-b}$? – Lutz Lehmann May 24 '14 at 12:48

1 Answers1

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Hint: The sum is a geometric series.

Robert Israel
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  • http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D0%7D%5E%7Bk%7D%282x%29%5Ei*%28x%2B1%29%5E%28k%2Bn-i%29&dataset= – math110 May 23 '14 at 07:19