Question

WILL MEDAL HELP!
Write out the Pk+1 statement for the following: 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6 . note: you need not prove this sentence to be true, just simply write out the statement.

Answer 1

where you see an \(n\) put a \(k+1\)

Answer 2

\[1^2+2^2+3^2+...+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} . \] then clean it up with some algebra

Answer 3

on the fraction , wouldnt you combine like terms, so \[1^2+2^2+3^2+...+(k+1)^2=\frac{ (2(k+1)+1) }{ 6 }\]

Answer 4

@satellite73

Answer 5

@satellite73 can you check my answer