I was just wondering if this equation:
"n^2 = n(n+1)(2n+1)/6"
is the equation for finding said question above.|||The formula should be 鈭?n^2 = n(n + 1)(2n + 1)/6, and is used to find the sum of the first n squares.|||The identity you have is if their was a Sigma sign
Sum of squares between 1 and n is
n(n+1)(2n+1)/6
Though what this has to do with the question as stated beats me
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