Geometric Progression sequence with an Arithmetic Progression grouping problem

[nicodemus1]

Good Day,

My friends and I are stuck on solving the last part of the attached problem.

The solution is 2^[(n^2 + n)/2] - 1.

Can anyone help us with solving this?

Thanks & Regards,
Nicodemus

 

[MarkFL]

The solution you give would be the sum of all the terms in the first n groups, not the sum of just the terms in the nth group.

Let:

$\displaystyle p<q$ where $\displaystyle p,q\in\mathbb{N}$

and then:

$\displaystyle S=2^p+2^{p+1}+2^{p+2}+\cdots+2^{q}$

$\displaystyle 2S=2^{p+1}+2^{p+2}+2^{p+3}+\cdots+2^{q}+2^{q+1}$

Subtracting the former from the latter, we find:

$\displaystyle S=2^{q+1}-2^p$

Now, let:

$\displaystyle p=\frac{n^2-n}{2},\,q=\frac{n^2+n}{2}-1$

 

[nicodemus1]

Good Day,

Thank you for the reply.

However, I don't see how it simplifies to the given solution. If it does, then I would first have to divide the expression by a term, right? How do I obtain that term and division from?

Thanks & Regards,
Nicodemus

 

[MarkFL]

The given solution is for:

$\displaystyle \sum_{k=0}^{\frac{n^2+n}{2}-1}2^k$

However, you are being asked to compute:

$\displaystyle \sum_{k=\frac{n^2-n}{2}}^{\frac{n^2+n}{2}-1}2^k$