SOLUTION: Prove that: {{{sum((2k-1^"")^""^""^"",k=1,n)}}}{{{""=""}}}{{{sum( ((n^2-1)/n^2)^(k-1),k=1,infinity)}}}

Algebra ->  Sequences-and-series -> SOLUTION: Prove that: {{{sum((2k-1^"")^""^""^"",k=1,n)}}}{{{""=""}}}{{{sum( ((n^2-1)/n^2)^(k-1),k=1,infinity)}}}      Log On


   

Question 1018961: Prove that: sum%28%282k-1%5E%22%22%29%5E%22%22%5E%22%22%5E%22%22%2Ck=1%2Cn%29%22%22=%22%22sum%28+%28%28n%5E2-1%29%2Fn%5E2%29%5E%28k-1%29%2Ck=1%2Cinfinity%29
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Prove that:
sum%28%282k-1%5E%22%22%29%5E%22%22%5E%22%22%5E%22%22%2Ck=1%2Cn%29%22%22=%22%22sum%28+%28%28n%5E2-1%29%2Fn%5E2%29%5E%28k-1%29%2Ck=1%2Cinfinity%29

The left side is the sum of the arithmetic series:

1 + 3 + 5 + ... + 2n-1

The formula for this sum is

S%5Bn%5D%22%22=%22%22expr%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29

where a1 = 1 and an = 2n-1 

S%5Bn%5D%22%22=%22%22expr%28n%2F2%29%281%2B%282n-1%29%5E%22%22%29%22%22=%22%22+expr%28n%2F2%29%281%2B2n-1%29%22%22=%22%22expr%28n%2F2%29%282n%29%22%22=%22%22+n%5E2

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Now we see if we can show that the right side is also 
equal to n2.


The right side is the sum of an infinite geometric series

The formula for this sum is 

S%5Binfinity%5D%22%22=%22%22a%5B1%5D%2F%281-r%29

where a1 = 1 and r+%22%22=%22%22+%28n%5E2-1%29%2Fn%5E2%22%22=%22%22n%5E2%2Fn%5E2-1%5E%22%22%2Fn%5E2%22%22=%22%221-1%5E%22%22%2Fn%5E2 

S%5Binfinity%5D%22%22=%22%221%5E%22%22%2F%281-%281-1%2Fn%5E2%29%29%22%22=%22%221%5E%22%22%2F%281-1%2B1%2Fn%5E2%29%22%22=%22%221%5E%22%22%2F%281%2Fn%5E2%29%22%22=%22%22n%5E2

So the right side also equals n2. 
That proves they are equal.

Edwin