Anonymous
Anonymous asked in Science & MathematicsEngineering · 8 years ago

Please solve by induction? I got close but ended up with an unbalanced term =(?

PBI: 1^2 + 2^2 + 3^2 + .... + n^2 = 1/6(n+1)(2n+1)

What i did was left the the assumed n=k as it was

and subbed it in for n = k+1 to LHS and RHS and expanded to show they were equal. Unless my algebra is wrong (and i'm pretty sure it always is), i am completely wrong lol.

Update:

I am sorry but it is

PBI: 1^2 + 2^2 + 3^2 + .... + n^2 = 1/6n(n+1)(2n+1)

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  • 8 years ago
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    To show it to be true for n = k + 1, you have the left hand side equal to

    1^2 + 2^2 + ... + k^2 + (k+1)^2

    Subbing your assumption for it being true for n = k gives

    k(k+1)(2k+1)/6 + (k+1)^2

    Then the algebra goes as follows:

    = [k(k+1)(2k+1)/6] + k² + 2k + 1

    = [(2k³+ 3k² + k)/6] + k² + 2k + 1

    = [(2k³+ 3k² + k)/6] + 6(k² + 2k + 1)/6

    = [(2k³+ 3k² + k)/6] + (6k² + 12k + 6)/6

    = [(2k³+ 3k² + k + 6k² + 12k + 6)/6

    = (2k³+ 9k² + 13k + 6)/6

    = (k+1)(k+2)(2k+3)/6.

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