Quickie #16

**Soroban** · January 27th 2007, 02:42 PM

Express the sum of the first $n$ terms of this sequence:

$1,\;(1+2),\;(1+2+2^2),\;(1+2+2^2+2^3),\;\cdots,\;( 1 + 2 + 2^2 + \cdots + 2^{n-1})$

**galactus** · January 27th 2007, 05:46 PM

The sum of the first n terms is $2^{n+1}-(n+2)$

For instance, sum of first term, n=1: $2^{2}-3=1$

Sum of first 2 terms: $2^{3}-4=4$

Sum of first 5 terms: $2^{6}-7=57$

The terms are $1, 3, 7, 15, 31, 63,....,2^{n-1}$

**anthmoo** · January 28th 2007, 01:30 AM

Originally Posted by galactus

The sum of the first n terms is $2^{n+1}-(n+2)$

For instance, sum of first term, n=1: $2^{2}-3=1$

Sum of first 2 terms: $2^{3}-4=4$

Sum of first 5 terms: $2^{6}-7=57$

The terms are $1, 3, 7, 15, 31, 63,....,2^{n-1}$

Are you sure?

For when n=3, ans=7

$2^{3-1}=2^2=4$ ??

Don't you mean...

$1,3,7,15,31,63,....,2^n-1$

*Edit: Oh! Sorry...you got it right at the first sentance.. >_<

**Soroban** · January 28th 2007, 07:24 AM

Good work, Galactus!

Here's my book's solution . . .

The $k^{th}$ term is: . $(1 + 2 + 2^2 + \cdots + 2^{k-1}) \:=\:2^k - 1$

The sum is: . $S_n\;=\;(2 - 1) + (2^2-1) + (2^3-1) + (2^4 - 1) + \cdots + (2^n-1)$

. . $S_n\;=\;\underbrace{(2 + 2^2 + 2^3 + \cdots + 2^n)}_{\text{geometric series}} - \underbrace{(1 + 1 + 1 + \cdots + 1)}_{n\text{ terms}}$

. . $S_n\;=\;2\,\frac{1-2^n}{1 - 2} - n\quad\Rightarrow\quad \boxed{S_n\;=\;2^{n+1} - 2 - n}$

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