using induction for a proof

[themathman]

This question is tripping me up and im not sure how to start it. the i^2 part is tripping me.

 

[Deleted member 4993]

This question is tripping me up and im not sure how to start it. the i^2 part is tripping me.

if you write out the summation for first 'n' terms, you get:

1² + 2² + 3² ........ + (n-2)² + (n-1)²+ n²

Continue....

 

[themathman]

if you write out the summation for first 'n' terms, you get:

1² + 2² + 3² ........ + (n-2)² + (n-1)²+ n²

Continue....

still confused as to how i should start the question with the n(n+1)/2 method

 

[MarkFL]

What you are given to prove is your induction statement \(P_k\). I would begin by verifying \(P_1\):

[MATH]\sum_{i=1}^1 i^2=\frac{1(1+1)(2(1)+1)}{6}[/MATH]
Once you verify that is true, then as your induction step, add \((k+1)^2\) to both sides:

[MATH]\sum_{i=1}^k i^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2[/MATH]
If you can show this is equivalent to:

[MATH]\sum_{i=1}^{k+1} i^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}[/MATH]
Then, you will have completed your proof by induction, because you will have derived \(P_{k+1}\) from \(P_k\).

 

[Steven G]

This question is tripping me up and im not sure how to start it. the i^2 part is tripping me.

first i =1, then i= 2, then i= 3,..., all the way up to i=k. Now you are adding up i²'s. So you get 1² + 2² + 3² + .... + k²

Please note that i² is NOT -1. i is NOT an imaginary number but rather the index which you are summing over