Question

What is the sum of all integral powers of 2 from 1 to 1050?

Geometric Series

Answer 1

We want to evaluate

`sum_(i=1)^(1050) 2^i`

We see that this is the sum of a geometric series with a ratio of 2. We can, therefore, use the well-known formula:
`S_n = (a_i(1-r^n))/(1-r)`

where `n` is the number of terms, `r` is the ratio, and `a_i` is the initial term. This means that we have
`S_1050 = (2(1-2^(1050)))/(1-2) = 2^(1051) - 2`

It is easy to see such a thing if we think of binary:
`2^i` is equivalent to `10...0` with `i` zeros. Therefore, the sum in binary is
`10 + 100 + ... = 111...110 = 10...000 - 10 = 2^(1051) - 2`

Answer 2

`:.S_1050=(2[2^1050-1])/(2-1)=2[2^1050-1]`

Explanation:

We know that,

`n^(th)term` of the Geometric series is :

`a_n=a_1(r)^(n-1)`

`where , a_1="first term and r=common ratio."`

Here ,

`S_n=2^1+2^2+2^3+...+2^1050`

`:.a_1=2^1 and r=2^2/2^1=4/2=2`

`a_n=2^1050=>n=1050`

`:." sum of first n-term is :"`

`S_n=(a_1(r^n-1))/(r-1)`

`:.S_1050=(2[2^1050-1])/(2-1)=2[2^1050-1]`