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anonymous

  • 4 years ago

prove by induction that 1^2 + 2^2 + 3^2 ... + n^2 = n(n+1)(2n+1)/6

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  1. phi
    • 4 years ago
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    Base case: n=1 1^2 =? 1*(1+1)*(2*1+1)/6 the left hand side is 1 the right hand side is 1*2*3/6 = 6/6 = 1. so n=1 works. Now assume the formula works for numbers up to n. Does it work for n+1? start with 1^2 + .... n^2 + (n+1)^2 now we assumed the formula works for the numbers up to n, so rewrite this sum as n(n+1)(2n+1)/6 + (n+1)^2 n(n+1)(2n+1)/6 + 6(n+1)(n+1)/6 [n(2n+1) + 6n+6](n+1)/6 [ 2n*n + 7n + 6](n+1)/6 (n+2)(2n+3)(n+1)/6 rename n+1 as m to make the answer more obvious note: (2n+3)= (n+n+1+1+1)= (n+1)+(n+1)+1= m+m+1= 2m+1 (m+1)(2m+1)m/6 which is the original formula

  2. anonymous
    • 4 years ago
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    so m = n + 1?

  3. phi
    • 4 years ago
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    yes, I only did that so we can compare to the original formula n(n+1)(2n+1)/6

  4. anonymous
    • 4 years ago
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    i see, Thanks alot!

  5. phi
    • 4 years ago
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    Turing test wrote it out nicely in the equation editor in your other post

  6. anonymous
    • 4 years ago
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    oh, ok

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