Method NO. 2 : Upseting Method, again


Lets calculate more compilated sum
S_n = \sum_{\substack{0 \leq k \leq n}}k2^k
further, S_n + (n+1)2^{n+1} = \sum_{\substack{0 \leq k \leq n}}(k+1)2^{k+1} the last one equals to \sum_{\substack{0 \leq k \leq n}}k2^{k+1} +\sum_{\substack{0 \leq k  \leq n}}2^{k+1} and now the first of the sums is equal to 2S_n however the second one is easy, cause its a geometric sequence equal to \frac{2-2^{n+2}}{1-2} = 2^{n+2} - 2 under the law from previous post. Now we have S_n + (n+1)2^{n+1} = 2S_n + 2^{n+2}-2 , in the result of elementary algebraic transformation we recieve S_n = (n-1)2^{n+1} + 2

It’s easy to notice that we could put everything in the place of 2, so let there be x. Then \sum_{k=0}^{n}kx^k = \frac{x-(n+1)x^{n+1}+nx^{n+2} }{(1-x)^2}

And now we have quite general law šŸ™‚


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