Induction proof for (n+1)^2 +(n+1)^2 +(n+1)^2 +...+(2n)^2 = n(2n+1)(7n+1)/6

[cameraman123]

1. Prove by mathematical induction, that (n+1)2 +(n+1)2 +(n+1)2 +...+(2n)2 = ((n)(2n+1)(7n+1))/6
   

 

[Deleted member 4993]

Prove by Induction that (n+1)^2 +(n+1)^2 +(n+1)^2 +...+(2n)^2 = n(2n+1)(7n+1)/6

Your post does not make sense!

Please put up the correct problem.

 

[pka]

1. Prove by mathematical induction, that (n+1)2 +(n+1)2 +(n+1)2 +...+(2n)2 = ((n)(2n+1)(7n+1))

I suspect it is \(\displaystyle \sum\limits_{k = 1}^n {{{(n + k)}^2}} = \frac{{n(2n + 1)(7n + 1)}}{6}\)

@cameraman, is that correct? If so think induction.

 

[cameraman123]

I suspect it is \(\displaystyle \sum\limits_{k = 1}^n {{{(n + k)}^2}} = \frac{{n(2n + 1)(7n + 1)}}{6}\)

@cameraman, is that correct? If so think induction.

Sorry about that! The correct problem is:
Prove By Mathematical Induction that:[h=2](n+1)^2 +(n+1)^2 +(n+1)^2 +...+(2n)^2 = n(2n+1)(7n+1)/6[/h]

 

[stapel]

Sorry about that! The correct problem is:
Prove By Mathematical Induction that:(n+1)^2 +(n+1)^2 +(n+1)^2 +...+(2n)^2 = n(2n+1)(7n+1)/6

Since we have no way of knowing how many copies of (n + 1)^2 occur, nor what happens in the "...", there is no way to proceed.

Please consult with your instructor regarding corrections of this exercise. Thank you!

 

[pka]

Sorry about that! The correct problem is:
Prove By Mathematical Induction thatn+1)^2 +(n+1)^2 +(n+1)^2 +...+(2n)^2 = n(2n+1)(7n+1)/6

I am absolutely sure that the problem I posted is the intended one. I have been around these questions long enough to recolonize it from the answer.
\(\displaystyle \begin{align*}\sum\limits_{k = 1}^n {{{\left( {n + k} \right)}^2}} &= \sum\limits_{k = 1}^n {\left( {{n^2} + 2nk + {k^2}} \right)}\\& = {n^3} + 2n\left( {\frac{{n(n + 1)}}{2}} \right) + \frac{{n(n + 1)(2n + 1)}}{6}\\&= \frac{{n(2n + 1)(7n + 1)}}{6}\end{align*}\)

SEE HERE

 

[lookagain]

1. Prove by mathematical induction, that (n+1)2 +(n+1)2 +(n+1)2 +...+(2n)2 = ((n)(2n+1)(7n+1))/6

cameraman123,

this is supposed to be the equivalent to your problem:


\(\displaystyle Prove \ \ by \ \ mathematical \ \ induction, \ \ for \ \ n \ \ge \ 1, \ \ that \)

\(\displaystyle (n + 1)^2 \ + \ (n + 2)^2 \ + \ (n + 3)^2 \ + \ ... \ + \ (2n)^2 \ = \ \dfrac{n(2n + 1)(7n + 1)}{6}.\)