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I did but i'm trying to prove it by substituting n with n+1 So far I have (1/6)(n+1)((n + 1) + 1)(2(n + 1) + 1) i'm trying to manipulate the parenthesis and whatnot to get it back to normal.
(1/6)(n+1)((n + 1) + 1)(2(n + 1) + 1) - (1/6)n(n + 1)(2n + 1) = (n+1)^2
my assumption of what needs to be done
i really dont see a way to nicely manipulate the parenthesis
(1/6)(n+1) [((n + 1) + 1)(2(n + 1) + 1) - n(2n + 1)] = (n+1)^2
(1/6) [((n + 1) + 1)(2(n + 1) + 1) - n(2n + 1)] = (n+1)
oh gosh im dreadfully afraid of getting it wrong at this point
(1/6) [((n + 2)(2n + 3) - (2n^2 + n)] = (n+1)
(1/6) [2n^2 +7n +6 - 2n^2 - n)] = (n+1)
haha ok I'm not sure you understood what i meant by "getting back to normal" I'm trying to get it where it goes back to the original (1/6)(n)(n + 1)(2n + 1)
1.) First we prove that the statement is true for n = 1: (1/6)(1)(2)(3) = 1. Yup so that works. (*) 2.) Now we prove that this is true for some n = k: 1^2 + 2^2+....+(k-1)^2 + k^2 = (1/6)(k)(k+1)(2k+1). Let's assume that this infact is true. (**) 3.) We now show that the statement remains true for some n = k + 1: 1^2 + 2^2+...+(k-1)^2 + k^2 + (k+1)^2 = (1/6)(k+1)(k+2)(2k+3) => 1^2 + 2^2+...+(k-1)^2 + k^2 = (1/6)(k+1)(k+2)(2k+3) - (k+1)^2 (***) Using our assumption that 1^2 + 2^2+....+(k-1)^2 + k^2 = (1/6)(k)(k+1)(2k+1), we substitute this for the left side in "(***)" and we try to show that the right hand side equals the left side: (1/6)(k)(k+1)(2k+1) = (1/6)(k+1)(k+2)(2k+3) - (k+1)^2 (****) Let's expand the right side to see if it looks like the right side: (1/6)(k+1)(k+2)(2k+3) - (k+1)^2 =(k+1)[(1/6)(k+2)(2k+3)-(k+1)] = (k+1)[(1/3)k^2 + (7/6)k + 1 - k - 1] = (k+1)[(1/3)k^2 + (1/6)k] = (k+1)(1/6)(k)(2k + 1) Clearly the answer is like our right hand side and so our assumption was correct. Hence the statement that 1^2+2^2+...+n^2 = (1/6)(n)(n+1)(2n+1) has now been proved. @iheartducks
i just got it! Thank you @genius12