Anonymous
Anonymous asked in Science & MathematicsEngineering · 8 years ago

# Please solve by induction? I got close but ended up with an unbalanced term =(?

PBI: 1^2 + 2^2 + 3^2 + .... + n^2 = 1/6(n+1)(2n+1)

What i did was left the the assumed n=k as it was

and subbed it in for n = k+1 to LHS and RHS and expanded to show they were equal. Unless my algebra is wrong (and i'm pretty sure it always is), i am completely wrong lol.

Update:

I am sorry but it is

PBI: 1^2 + 2^2 + 3^2 + .... + n^2 = 1/6n(n+1)(2n+1)

Relevance
• 8 years ago

To show it to be true for n = k + 1, you have the left hand side equal to

1^2 + 2^2 + ... + k^2 + (k+1)^2

Subbing your assumption for it being true for n = k gives

k(k+1)(2k+1)/6 + (k+1)^2

Then the algebra goes as follows:

= [k(k+1)(2k+1)/6] + k² + 2k + 1

= [(2k³+ 3k² + k)/6] + k² + 2k + 1

= [(2k³+ 3k² + k)/6] + 6(k² + 2k + 1)/6

= [(2k³+ 3k² + k)/6] + (6k² + 12k + 6)/6

= [(2k³+ 3k² + k + 6k² + 12k + 6)/6

= (2k³+ 9k² + 13k + 6)/6

= (k+1)(k+2)(2k+3)/6.