Question

A statement $S_n$ about the positive integers is given. Write statements $S_k$ and $S_{k+1}$, simplifying $S_{k+1}$ completely. $S_n$: $1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2 - 3n - 1)}{2}$

          A statement $S_n$ about the positive integers is given. Write statements $S_k$ and $S_{k+1}$, simplifying $S_{k+1}$ completely.
$S_n$: $1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2 - 3n - 1)}{2}$

A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely.
Sn: 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = (n(6n^2 - 3n - 1))/(2)

Added by Anna M.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Ankit Gupta

06/05/2020

Step 1

$S_k$: $1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \frac{k(6k^2 - 3k - 1)}{2}$ Show more…

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A statement $S_n$ about the positive integers is given. Write statements $S_k$ and $S_{k+1}$, simplifying $S_{k+1}$ completely. $S_n$: $1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2 - 3n - 1)}{2}$