Proving Induction for Summation of Squares: n ≥ 1 | Homework Solutions

[dlb89]

Homework Statement

Prove that for n $$\geq$$ 1
n
$$\sum$$ m$$^{2}$$ = (1/6)*n(n+1)(2n+1)
m=1

Homework Equations

The Attempt at a Solution

Base Case:
n=1

(1)^2 = (1/6)*(6)
1=1

Inductive Step:
Assume n=k
k
$$\sum$$ m^2 = (1/6)*k(k+1)(2k+1) is true
m=1


Let n=k+1

k+1
$$\sum$$ m^2 = (1/6)*(k+1)((k+1)+1)(2(k+1)+1)
m=1

Any hints to prove this? I've tried simplifying the last equation but I haven't had any luck

 

[Mark44]

Split your last summation into one running from m = 1 to k plus the single term when m = k + 1. For the first summation, use what you have from your induction step.

 

[HallsofIvy]

Homework Statement

Prove that for n $$\geq$$ 1
n
$$\sum$$ m$$^{2}$$ = (1/6)*n(n+1)(2n+1)
m=1

Homework Equations

The Attempt at a Solution

Base Case:
n=1

(1)^2 = (1/6)*(6)
1=1

Inductive Step:
Assume n=k
k
$$\sum$$ m^2 = (1/6)*k(k+1)(2k+1) is true
m=1


Let n=k+1

k+1
$$\sum$$ m^2 = (1/6)*(k+1)((k+1)+1)(2(k+1)+1)
m=1

Well, this is what you want to get.

Any hints to prove this? I've tried simplifying the last equation but I haven't had any luck

Recognise that
$$\sum_{m=0}^{k+1} m^2= \sum_{m=0}^k m^2+ (k+1)^2$$

If
$$\sum_{m=0}^k m^2= (1/6)*k(k+1)(2k+1)$$
then that becomes
$$\sum_{m=0}^{k+1} m^2= (1/6)*k(k+1)(2k+1)+ (k+1)^2$$
That is what you must show is equal to
$$(1/6)*(k+1)((k+1)+1)(2(k+1)+1)$$