mathematical induction?

Question

prove that 1^2 + 2 ^2 + 3 ^2 + ....... + n^2 = {n(n+1)(2n+1)} / 6 by Mathematical induction.

Answer 1

1^2 + 2 ^2 + 3 ^2 + ....... + n^2 = n(n+1)(2n+1) / 6 ----------------- equation 1

let P(n) : 1^2 + 2 ^2 + 3 ^2 + ....... + n^2 = n(n+1)(2n+1) / 6
P(1) = 1^2 = 1 (1 + 1) ( 2 * 1 + 1) / 6 = 2 * 3 / 6 = 6 / 6 = 1

P( 1 ) is true because LHS = RHS

Assume P (m) : 1^2 + 2^2 + ......... m^ 2 = m(m+1)(2m+1) / 6

add the next term ( m +1) ^2 on both sides, 1 ^2 + 2 ^2 + .......... = m ^2 + (m + 1 ) ^ 2
= m(m+1)(2m + 1) / 6 + ( m + 1) ^2
= ( m + 1) [ m ( 2m + 1) / 6 + ( m + 1 ) ]
= ( m + 1 ) [ {m ( 2m + 1 ) + 6 ( m + 1)} / 6 ]
= (m + 1) [ {2m^2 + m + 6m + 6 } / 6
= ( m + 1) [{ 2m ^2 + 7m + 6 }/6]
= ( m + 1 ) [ {2m ^ 2 + 4m + 3 m + 6 }/ 6 ]
= ( m + 1) [{ 2m ( m + 2) + 3 ( m + 2 ) } / 6 ]
= (m + 1) [ { (m + 2 ) ( 2m + 3 ) } / 6 ]
= [ ( m + 1) { (m + 1) + 1} { 2 ( m + 1) + 1 } ] / 6
in the above obtained equation m + 1 is in place of n of the equation 1
therefore the equation is P (m + 1)

Thus P(m) => P ( m +1)

By the principle of mathematical induction, P (n) is true for all the values of n belonging to N.

Answer 2

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

1) Base case: Let n = 1. Then

LHS = 1^2 ... 1^2
= 1 ... 1
= 1

RHS = 1(1 + 1)(2*1 + 1)/6
= 1(2)(3)/6
= 6/6
= 1

LHS = RHS, so the formula holds true for n = 1.

2) Induction Hypothesis: Assume that the formula holds true for up to n = k for some value k. That is,

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

(We want to prove that
1^2 + 2^2 + 3^2 + ... + (k + 1)^2 = (k + 1)(k + 1 + 1)(2(k + 1) + 1)/6 )

But, the sum of the first (k + 1) terms is equal to the sum of k terms, plus the (k + 1) term. i.e.

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

By our induction hypothesis, 1^2 + ... + k^2 = k(k + 1)(2k + 1)/6, so we have

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

Factoring out (1/6)(k + 1), we get

= (1/6)(k + 1) [ k(2k + 1) + 6(k + 1) ]
= (1/6)(k + 1) [ 2k^2 + k + 6k + 6 ]
= (1/6)(k + 1) [ 2k^2 + 7k + 6 ]
= (1/6)(k + 1) (2k + 3)(k + 2)
= (1/6)(k + 1)(k + 2)(2(k + 1) + 1)
= (1/6)(k + 1)(k + 1 + 1)(2(k + 1) + 1)
= (k + 1)(k + 1 + 1)(2(k + 1) + 1)/6

Which shows the formula holds true for n = k + 1.

Therefore, by the principle of mathematical induction,

1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6
for all natural numbers n.

Answer 3

P(n): 1² + 2² + -------n² = n.(n + 1).(2n + 1) / 6
Let k be a value of n.
P(k): 1² + 2² + ----k² = k.(k + 1).(2k + 1) / 6
Have now to prove P(1) true and P(k + 1) true.
Consider P(1)
LHS = 1² = 1
RHS = (1 x 2 x 3)/6 = 1
Thus P(1) is true.
Consider P(k + 1)
P(k + 1):
1² + 2² -----(k + 1)² = (k + 1).(k + 2).(k + 3) / 6
Have now to prove this to be the case.
P(k): 1² + 2² +----k² = k.(k + 1).(2k + 1) / 6 is true.
1²+ 2² --k² + (k + 1)² = k.(k + 1).(2k + 1)/6 +(k + 1)²
RHS = [ k.(k + 1).(2k + 1) + 6(k + 1)² ] / 6
RHS = (k + 1).[k.(2k + 1) + 6(k + 1)] / 6
RHS = (k + 1).[ 2k² + 7k + 6 ] / 6
RHS = (k + 1).(k + 2).(2k + 3) / 6
Thus P(k + 1) is true.
P(k) is true
P(1) is true
P(k + 1) is true
THUS P(n) is true