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

Question

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?

Answer 1

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 $$ \begin{align} (2^0+\cdots+2^n)\times(1+2+2^n) &=(2^{n+1}-1)\times(2^n+3) \\&=2^{2n+1}+3\times2^{n+1}-2^n-3. \\&=2^{2n+1}+5\times2^n-3. \end{align} $$

Answer 2

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

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

Answer 3

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

Answer 4

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 $$S=\sum_{i=0}^n S_i=\left(2^{n+1}-1\right)^2$$

Answer 5

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