- #1
mxam
- 10
- 0
Can you help me with this exercise?
[itex]1^{1}[/itex]+[itex]2^{2}[/itex]+[itex]3^{3}[/itex]+[itex]4^{4}[/itex]+...+[itex]n^{n}[/itex] = [itex]n^{n+1}[/itex]
Thanks!
PD. I was trying to solve, and i have this:
[itex]1^{1}[/itex]=[itex]1^{1+1}[/itex] =
1 = 1
a) [itex]k^{k+1}[/itex]
b) [itex]k+1^{k+1}[/itex] = [itex]k+1^{(k+1)+1}[/itex]
a in b) [itex]k^{k+1}[/itex] + [itex]k+1^{k+1}[/itex] = [itex]k+1^{(k+1)+1}[/itex]
([itex]k)^{k}[/itex]([itex]k)^{1}[/itex]+([itex]k+1)^{k}[/itex]([itex]k+1)^{1}[/itex]=([itex]k+1)^{k}[/itex]([itex]k+1)^{1}[/itex]([itex]k+1)^{1}[/itex]
I´m lost in this step . . . Thanks again!
[itex]1^{1}[/itex]+[itex]2^{2}[/itex]+[itex]3^{3}[/itex]+[itex]4^{4}[/itex]+...+[itex]n^{n}[/itex] = [itex]n^{n+1}[/itex]
Thanks!
PD. I was trying to solve, and i have this:
[itex]1^{1}[/itex]=[itex]1^{1+1}[/itex] =
1 = 1
a) [itex]k^{k+1}[/itex]
b) [itex]k+1^{k+1}[/itex] = [itex]k+1^{(k+1)+1}[/itex]
a in b) [itex]k^{k+1}[/itex] + [itex]k+1^{k+1}[/itex] = [itex]k+1^{(k+1)+1}[/itex]
([itex]k)^{k}[/itex]([itex]k)^{1}[/itex]+([itex]k+1)^{k}[/itex]([itex]k+1)^{1}[/itex]=([itex]k+1)^{k}[/itex]([itex]k+1)^{1}[/itex]([itex]k+1)^{1}[/itex]
I´m lost in this step . . . Thanks again!
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