Question

Prove that each statement is true for all positive integers. What are the fist few steps?
1 * 3 + 3 * 5 + 5 * 7 + ... + (2n-1)(2n+1) = n(4n^2 + 6n - 1) / 3

Answer 1

induction proof?

Answer 2

what kind of proof are you needing? i think you have to do it by induction

Answer 3

\[\sum _{k=1}^n \left(4
k^2-1\right)= \frac{1}{3} n \left(4 n^2+6
n-1\right)

\]
Notice that it is true for n=1.
Suppose that it is true for for the sum up to n
and prove it for the sum up to n+1
which means suppose that
\[\sum _{k=1}^n \left(4
k^2-1\right)= \frac{1}{3} n \left(4 n^2+6
n-1\right)

\]
and show that
\[\sum _{k=1}^{n+1} \left(4
k^2-1\right)= \frac{1}{3} (n+1) \left(4 (n+1)^2+6
( n+1)-1\right)

\]\\

\]