Jan 8, 2013 Thread starter #1 N nicodemus New member Jul 22, 2012 16 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 any one help us with solving this? Thanks & Regards, Nicodemus
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 any one help us with solving this? Thanks & Regards, Nicodemus
Jan 8, 2013 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 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: $\displaystyle p=\frac{n^2-n}{2},\,q=\frac{n^2+n}{2}-1$
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: $\displaystyle p=\frac{n^2-n}{2},\,q=\frac{n^2+n}{2}-1$
Jan 9, 2013 Thread starter #3 N nicodemus New member Jul 22, 2012 16 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
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
Jan 9, 2013 Admin #4 M MarkFL Administrator Staff member Feb 24, 2012 13,775 The given solution is for: $\displaystyle \sum_{k=0}^{\frac{n^2+n}{2}-1}2^k$ However, you are being asked to compute: $\displaystyle \sum_{k=\frac{n^2-n}{2}}^{\frac{n^2+n}{2}-1}2^k$
The given solution is for: $\displaystyle \sum_{k=0}^{\frac{n^2+n}{2}-1}2^k$ However, you are being asked to compute: $\displaystyle \sum_{k=\frac{n^2-n}{2}}^{\frac{n^2+n}{2}-1}2^k$