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F(n)= 7+n+f(n-1), where f(0)=1 Can anyone help with this one? :)?

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  • F(n)= 7+n+f(n-1), where f(0)=1 Can anyone help with this one? :)?


Answers

Answer #1 | 10/05 2016 10:31
Using your rule f(1) = 7 + 1 + 1 = 9 f(2) = 7 + 2 + 9 = 18 f(3) = 7 + 3 + 18 = 28 . . .
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Answer #11 | 10/05 2016 10:35
Well, @unknonwn : thank you : corrected let n = 2p+1 (or : p = (n - 1)/2 ) then 1^2+3^2+5^2+7^2+.....+n^2 = Σ(k from 0 to p) (2k + 1)^2 = Σ(k from 0 to p) [ 4k^2 + 4k + 1 ] = 4 Σ(k from 0 to p) k^2 + 4 Σ(k from 0 to p) k + Σ(k from 0 to p) 1 = 4 p(p+1)(2p+1)/6 + 4 p(p+1)/2 + p + 1 hope it' ll help !!
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Answer #12 | 10/05 2016 10:42
1² + 3² + 5² + 7² + ... + n² = 1² + 2² + 3² + .... + n² - 2² - 4² - 6² - ... - (n-1)² = n(n+1)(2n+1)/6 - 2²(1² + 2² + ... + ((n-1)/2)²) = n(n+1)(2n+1)/6 - 4 ((n-1)/2)((n-1)/2+1)(n)/6 = (n/6) ((n+1)(2n+1) - (n-1)(n+1)) = n(n+1)(n+2)/6 ( = ((n+2)!/(n-1)!)/3! = C(n+2,3) ) Michael : Sum(k from 0 to p) 1 = p + 1 and not p
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Answer #13 | 10/05 2016 03:42
1² + 3² + 5² + 7² + ... + n² = 1² + 2² + 3² + .... + n² - 2² - 4² - 6² - ... - (n-1)² = n(n+1)(2n+1)/6 - 2²(1² + 2² + ... + ((n-1)/2)²) = n(n+1)(2n+1)/6 - 4 ((n-1)/2)((n-1)/2+1)(n)/6 = (n/6) ((n+1)(2n+1) - (n-1)(n+1)) = n(n+1)(n+2)/6 ( = ((n+2)!/(n-1)!)/3! = C(n+2,3) ) Michael : Sum(k from 0 to p) 1 = p + 1 and not p
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Answer #14 | 10/05 2016 03:35
Well, @unknonwn : thank you : corrected let n = 2p+1 (or : p = (n - 1)/2 ) then 1^2+3^2+5^2+7^2+.....+n^2 = Σ(k from 0 to p) (2k + 1)^2 = Σ(k from 0 to p) [ 4k^2 + 4k + 1 ] = 4 Σ(k from 0 to p) k^2 + 4 Σ(k from 0 to p) k + Σ(k from 0 to p) 1 = 4 p(p+1)(2p+1)/6 + 4 p(p+1)/2 + p + 1 hope it' ll help !!
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