Question

Find the sum of 3+7+14+27+52 +...... to n terms?

Answer 1

`3*2^n+1/2n^2-1/2n-3`

Explanation:

Given:

`3+7+14+27+52+...`

Note that no finite initial sequence determines the following terms without further information. So it could continue in any way.

That having been said, we can try to match the given terms with some kind of regular formula.

We could choose to match the given terms with a polynomial formula, which would be one possibility, but let us look for a simpler pattern.

Note that the terms of the sequence roughly double each time, so let's try subtracting the geometric progression:

`3, 6, 12, 24, 48,...`

The result is:

`0, 1, 2, 3, 4,...`

So we can write a formula for the `n`th term as:

`a_n = 3*2^(n-1)+n-1`

The sum to `N` terms is the sum of a geometric progression and an arithmetic one.

For geometric progressions we have:

`sum_(n=1)^N ar^(n-1) = (a(r^N-1))/(r-1)`

and for arithmetic progressions we have:

`sum_(n=1)^N a+d(n-1) = aN+1/2dN(N-1)`

So

`sum_(n=1)^N 3*2^(n-1)+n-1 = (3(2^N-1))/(2-1)+0(N)+1/2(1)N(N-1)`

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

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