Question #9058

Prove by principle of Mathematical induction that sum of squares of first n natural
number is {n(n+1)(2n+1)}/6

Expert's answer

Let n=1

Then

{n(n+1)(2n+1)}/6 = 1*2*3/6 = 1

and

1^2=1

as well.

Suppose that we have proved that

1^2 + 2^2 + ...

+ k^2 = {k(k+1)(2k+1)}/6

for all k<n+1

Let us prove this identity for

k=n+1, that is

1^2 + 2^2 + ... + n^2 + (n+1)^2 = {(n+1)(n+2)(2(n+1)+1)}/6

= {(n+1)(n+2)(2n+3)}/6

Consider

the sum

S = 1^2 + 2^2 + ... + n^2 + (n+1)^2 ** by inductive

step

= {n(n+1)(2n+1)}/6 + (n+1)^2

= [ n(2n+1) +

6(n+1) ] * (n+1)/6

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

= [

2n^2 + 7n + 6 ] * (n+1)/6

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

=

{(n+1)(n+2)(2n+3)}/6

Then

{n(n+1)(2n+1)}/6 = 1*2*3/6 = 1

and

1^2=1

as well.

Suppose that we have proved that

1^2 + 2^2 + ...

+ k^2 = {k(k+1)(2k+1)}/6

for all k<n+1

Let us prove this identity for

k=n+1, that is

1^2 + 2^2 + ... + n^2 + (n+1)^2 = {(n+1)(n+2)(2(n+1)+1)}/6

= {(n+1)(n+2)(2n+3)}/6

Consider

the sum

S = 1^2 + 2^2 + ... + n^2 + (n+1)^2 ** by inductive

step

= {n(n+1)(2n+1)}/6 + (n+1)^2

= [ n(2n+1) +

6(n+1) ] * (n+1)/6

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

= [

2n^2 + 7n + 6 ] * (n+1)/6

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

=

{(n+1)(n+2)(2n+3)}/6

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