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Is there any formula for the series below: $$ (2^0)n + (2^1)(n-1) + (2^2)(n-2) + (2^3)(n-3) + \cdots +(2^{n-1})(1) $$ If no, please let me know how to solve such series

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3 Answers

Here are some tools to solve your problem.

  • The first basic observation, going back to Zeno, is that $\sum\limits_{k=1}^n2^{-k}=1-2^{-n}$.

  • More generally, for every $x$, $S_n(x)=\sum\limits_{k=0}^{n-1}x^{k}=\frac{x^n-1}{x-1}$ (watch out for the shift of indexation).

  • Differentiating this with respect to $x$, one gets a formula for $T_n(x)=\sum\limits_{k=0}^{n-1}nx^{n}.$

Now, you are interested in the sum $U_n(x)=\sum\limits_{k=0}^{n-1}x^k(n-k)$ (which is not a series, by the way) for $x=2$. Your first task is to express $U_n(x)$ in terms of some $S_n(x)$ and $T_n(x)$ defined above.

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Due to the symmetry, rewrite the sum as

$$\sum_{k=0}^{n} k \, 2^{n-k} = 2^n \sum_{k=0}^{n} k \, \left ( \frac{1}{2} \right )^k $$

You may use the fact that

$$\sum_{k=0}^{n} k \, r^k = \frac{r-r^{n+1}}{(1-r)^2} - n \frac{r^{n+1}}{1-r} $$

with $r=1/2$. The result is

$$2^{n+1}-2-n$$

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If you meant $$2^0(n)+2^1(n-1)+2^2(n-2)+\cdots+2^{n-2}(2)+2^{n-1}(1)$$

Let $$S=2^0(n)+2^1(n-1)+2^2(n-2)+\cdots+2^{n-2}(2)+2^{n-1}(1)$$

So, $$2S=2^1(n)+2^2(n-1)+2^3(n-2)+\cdots+2^{n-1}(2)+2^{n}(1)$$

So, $$2S-S=2^1\{n-(n-1)\}+2^2\{(n-1)-(n-2)\}\cdots+2^{n-2}(3-2)+2^{n-1}(2-1)+2^n-n$$

$$S=2^1+2^2+\cdots+2^{n-1}+2^n-n=\frac{2(2^{n}-1)}{2-1}-n=2^{n+1}-n-2$$

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