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Exact closed form of this expression $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$

I assume this means there is just one $2^0$ and one $2^{2n}$ and a double of all the terms in between?

Benjamin
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    Not quite, its a bit more complicated than that. The $n^{th}$ power appears $n+1$ times. Never mind, I misunderstood your question, there are no dots as in $+\cdots+$. – ReverseFlowControl Nov 15 '14 at 05:53
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    Is it $(2^0+…+2^n)+(2^1+…+2^{n+1})+(2^n+…+2^{2n})$ or $(2^0+…+2^n)+(2^1+…+2^{n+1})+\cdots +(2^n+…+2^{2n})$ ? Please clarify. – Claude Leibovici Nov 15 '14 at 06:12
  • Sorry guys this question was sent to me by my friend, and I asked for his original printed pdf to check. He mistyped the question and the original question was indeed the one mentioned by @Claude Leibovici. Sorry for the confusion caused. I shall now edit the question back. – Benjamin Nov 15 '14 at 06:33

5 Answers5

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Hint: Each of the brackets has a common factor of $(2^0+2^1+...+2^n)$

Empy2
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Your series is

$1+2+\cdots+2^n$

$\,\,\,\,\,\,\,\,\,2+\cdots+2^n+2^{n+1}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2^n+2^{n+1}+\cdots+2^{2n}$

And Sum can be written as

$$\begin{align}s&=1+2(2+4+\cdots+2^{n})+2^{n+1}+2^n+2^{n+1}+2^{n+2}+\cdots2^{2n}\\ &=1+2(2+4+\cdots+2^{n})+2^{n+1}+(2^n+2^{n+1}+2^{n+2}+\cdots2^{2n})\\ \end{align}$$

So, answer is

$$1+2\cdot\frac{2^{n+1}-2}{2-1}+2^{n+1}+\frac{2^{2n+1}-2^{n}}{2-1}=2^{2n+1}+3\cdot2^{n+1}-2^n-3$$

Aditya Hase
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Assuming that the problem is $$(2^0+…+2^n)+(2^1+…+2^{n+1})+\cdots +(2^n+…+2^{2n})$$ and not $$(2^0+…+2^n)+(2^1+…+2^{n+1})+(2^n+…+2^{2n})$$

consider the general terms in brackets $$S_i=\sum_{k=i}^{n+i}2^k$$ As already said, it is a geometric progression and the classical formula leads then to $$S_i=2^i \left(2^{n+1}-1\right)$$ which is again geometric progression and then $$S=\sum_{i=0}^n S_i=\left(2^{n+1}-1\right)^2$$

Claude Leibovici
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  • If you read the question carefully there are only three terms in the sum...that is three bracketed terms. Unless the OP meant something else. – ReverseFlowControl Nov 15 '14 at 05:57
  • @Genomeme. Oooops ! I thought that there were $\cdots$. I sall ask for clarification and eventually delete my answer. Thanks for pointing out. – Claude Leibovici Nov 15 '14 at 06:11
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Close: there would be 3 copies of $2^n$, 1 copy of each $2^0$ and $2^{n+2}$ through $2^{2n}$, and 2 copies of everything else

enter image description here

Simpson17866
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Since all terms are divisible by $2^0+\cdots+2^n$, you can split off that common factor: $$ (2^0+\cdots+2^n)\times(2^0+\cdots+2^n)=(2^{n+1}-1)^2=2^{2n+2}-2^{n+2}+1. $$

In fact, the above misreads the question to be about $(2^0+\cdots+2^n)+(2^1+\cdots+2^{n+1})+\cdots+(2^n+\cdots+2^{2n})$ rather than about $(2^0+\cdots+2^n)+(2^1+\cdots+2^{n+1})+(2^n+\cdots+2^{2n})$ as it actually states (in title and body). The latter question is less interesting, so I'll assume that OP actually meant the former with the extra ellipses and leave the answer above. But just in case a sum of precisely $3$ parenthesised sum is meant; that gives $$ \begin{align} (2^0+\cdots+2^n)\times(1+2+2^n) &=(2^{n+1}-1)\times(2^n+3) \\&=2^{2n+1}+3\times2^{n+1}-2^n-3. \\&=2^{2n+1}+5\times2^n-3. \end{align} $$

Marc van Leeuwen
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  • sorry but I do not understand where did the multiply come from in your expression? – Benjamin Nov 15 '14 at 05:57
  • I've made it more explicit. – Marc van Leeuwen Nov 15 '14 at 06:02
  • why does the second term $(2^1+...+2^{n+1})$ has a factor of $2^0+...+2^n$? Shouldn't it only has a factor of $2^1+...+2^n$? – Benjamin Nov 15 '14 at 06:12
  • Benjamin...before you even attempt to understand this answer, you must be clear about your original question. That is, did you mean to say there are exactly 3 terms in parenthesis? OR, did you mean there is a series of $n$, or $n+1$ as the case may be, such terms in parenthesis? If you mean the question as it is currently stated, then this is not the answer you are looking for. – ReverseFlowControl Nov 15 '14 at 06:17
  • @Benjamin No. $2^1+\cdots+2^{n+1}=2\times(2^0+\cdots+2^n)$. – Marc van Leeuwen Nov 15 '14 at 06:17