How to compute 'z', where $\displaystyle z = \sum_{n=1}^{100} n\times 2^n$ ?
The answer is of the form $99 \times 2^{101} + 2$, I need a fast approach as this problem is supposed to be solved under a minute.
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Martin Sleziak
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Quixotic
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Hint: start from the binomial theorem and differentiate accordingly... – J. M. ain't a mathematician Jul 24 '11 at 16:20
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@Aryabhata:Thanks for that pointing out to that question. – Quixotic Jul 24 '11 at 20:31
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If you define $y=\sum_{n=1}^{100}x^n=\frac{x^{101}-1}{x-1}$, then $z=x\frac{dy}{dx}$ evaluated at $x=2$. I love taking the derivative with respect to $2$.
Ross Millikan
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4Yes, lotta fun differentiating with respect to constants, but best to do it before they completely settle down. :) – paul garrett Jul 24 '11 at 20:01
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4Failing to differentiate between variables and constants invariably and constantly leads to trouble. – joriki Jul 24 '11 at 21:08
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We can actually solve this problem without calculus. (See this answer.)
Full Solution: Lets find a general formula for the following sum for any $r,m$: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$
This can be derived in a similar manner to the formula for the geometric series. Notice that $$S_{m}-rS_{m}=-mr^{m+1}+\sum_{n=1}^{m}r^{n}$$
$$=-mr^{m+1}+\frac{r-r^{m+1}}{1-r}=\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}.$$ Hence $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r$, so by letting $r=2$ we are able to conclude
$$\sum_{n=1}^m n2^n = m2^{m+2}-(m+1)2^{m+1}+2=2^{m+1}(m-1)+2.$$
Hope that helps,
Eric Naslund
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