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:

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

 

[CellCoree]

Hello Nicodemus,

I would be happy to help you with solving this problem. The first step is to understand the given solution. The term 2^[(n^2 + n)/2] represents a geometric progression with a common ratio of 2. This means that each term in the sequence is multiplied by 2 to get the next term. The exponent, (n^2 + n)/2, is the number of terms in the sequence.

The term -1 at the end represents an arithmetic progression with a common difference of -1. This means that each term in the sequence is subtracted by 1 to get the next term. The -1 at the end is subtracted to account for the first term in the arithmetic progression, which is 1.

To solve the problem, you need to first find the number of terms in the geometric progression, which is (n^2 + n)/2. Then, you need to find the sum of the geometric progression, which is given by the formula (a * (r^n - 1))/(r-1), where a is the first term and r is the common ratio. In this case, a = 1 and r = 2.

Once you have the sum of the geometric progression, you can add the -1 at the end to account for the first term in the arithmetic progression. This will give you the final solution of 2^[(n^2 + n)/2] - 1.

I hope this helps you understand the solution and solve the problem. Let me know if you need any further clarification or assistance.

Best regards,