- #1
AGNuke
Gold Member
- 455
- 9
If [itex]f(r) = 1 + \frac{1}{2} + \frac{1}{3}+... \frac{1}{r}[/itex] and [itex]f(0) = 0[/itex], then find [itex]\sum_{r=1}^{n}(2r+1)f(r)[/itex]
I tried the following steps but stuck at some point.
[itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r)[/itex]
[itex]= \sum_{r=1}^{n}rf(r)+(r+1)[f(r+1)-\frac{1}{r+1}][/itex]
[itex]= \sum_{r=1}^{n}rf(r)+(r+1)f(r+1) - \sum_{r=1}^{n}1[/itex]
The real problem is to find [itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r+1)[/itex]
Can someone please help me beyond this point by solving the above steps?
For verification, the Solution is [itex](n+1)^{2}f(n+1)-\frac{n^2+3n+2}{2}[/itex]
Thanks in Advanced!
I tried the following steps but stuck at some point.
[itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r)[/itex]
[itex]= \sum_{r=1}^{n}rf(r)+(r+1)[f(r+1)-\frac{1}{r+1}][/itex]
[itex]= \sum_{r=1}^{n}rf(r)+(r+1)f(r+1) - \sum_{r=1}^{n}1[/itex]
The real problem is to find [itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r+1)[/itex]
Can someone please help me beyond this point by solving the above steps?
For verification, the Solution is [itex](n+1)^{2}f(n+1)-\frac{n^2+3n+2}{2}[/itex]
Thanks in Advanced!
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