Question

Problem 2
Prove that
1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 6
For all positive integers n.

Answer 1

Answer:

Step-by-step explanation:

Statement P (n) is defined by  

1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 2  

STEP 1: We first show that p (1) is true.  

Left Side = 1 2 = 1  

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1  

Both sides of the statement are equal hence p (1) is true.  

STEP 2: We now assume that p (k) is true  

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

and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement  

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

Set common denominator and factor k + 1 on the right side  

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6  

Expand k (2k + 1)+ 6 (k + 1)  

= (k + 1) [ 2k 2 + 7k + 6 ] /6  

Now factor 2k 2 + 7k + 6.  

= (k + 1) [ (k + 2) (2k + 3) ] /6  

We have started from the statement P(k) and have shown that  

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

Which is the statement P(k + 1).

Answer 2

Prove that 12 + 22 + ... + n2 = [n(n+1)(2n+1)]/6, for every positive integer n.

Proof:

I. Prove that the equation holds for n = 1

If n = 1, then [n(n+1)(2n+1)]/6 = [(1)(2)(3)]/6 = 1.
So, 12 = [(1)(1+1)(2(1)+1)]/6

II. Assume that the equation holds for n and prove that the equation holds for n+1.

Assume: 12 + 22 + ... + n2 = [n(n+1)(2n+1)]/6

Prove: 12 + 22 + ... + n2 +(n+1)2 =
[(n+1)(n+2)(2n+3)]/6

By assumption, 12 + 22 +... + n2 + (n+1)2 =

[n(n+1)(2n+1)]/6 + (n+1)2

= [n(n+1)(2n+1) +6(n+1)2]/6

= [(n+1){n(2n+1) + 6(n+1)}]/6

=[(n+1){2n2 + 7n +6}]/6

= [(n+1)(n+2)(2n+3)]/6