Question

`9+17+32+61+118+231+456+---+n=?`

Answer 1

The sum to `N` terms is:

`7*2^N+1/2N^2+3/2N-7`

Explanation:

Given:

`9+17+32+61+118+231+456+...`

It looks like the `n`th term is defined recursively by:

`{ (a_1 = 9), (a_(n+1) = 2a_n - n " for " n >= 1) :}`

Consider the sequence defined by `7*2^(n-1)`. It's first few terms are:

`7, 14, 28, 56, 112, 224, 448,...`

Subtracting these from the given sequence, we get the sequence:

`2, 3, 4, 5, 6, 7, 8,...`

So it looks like we can assume the general formula:

`a_n = 7*2^(n-1)+n+1`

The general term of a geometric series can be written:

`b_n = b r^(n-1)`

where `b` is the initial term and `r` the common ratio.

Its sum to `N` terms is:

`(b(r^N-1))/(r-1)`

Let `b = 7` and `r = 2`.

Then we find:

`sum_(n=1)^N 7*2^(n-1) = 7(2^N - 1)`

Note that:

`sum_(n=1)^N n = 1/2 N(N+1)`

Also:

`sum_(n=1)^N 1 = N`

So:

`sum_(n=1)^N a_n = sum_(n=1)^N (7*2^(n-1)+n+1)`

`color(white)(sum_(n=1)^N a_n) = sum_(n=1)^N 7*2^(n-1)+sum_(n=1)^N n+sum_(n=1)^N 1`

`color(white)(sum_(n=1)^N a_n) = 7(2^N-1)+1/2N(N+1)+N`

`color(white)(sum_(n=1)^N a_n) = 7*2^N-7+1/2N^2+1/2N+N`

`color(white)(sum_(n=1)^N a_n) = 7*2^N+1/2N^2+3/2N-7`