Question

Using mathematical induction, prove that 1² + 2² + 3² + ... + n² = $\frac{n(n + 1)(2n + 1)}{6}$ for all n ∈ N

Answer 1

HELLO DEAR,




let p(n) = 1² + 2² + 3² + 4² + ....... n² = 1/6[n(n + 1)(2n + 1)]

for n = 1,

L.H.S = 1² = 1

R.H.S = 1/6[1(1 + 1)(2 + 1)] = 1/6[2 * 3] = 1


Hence, L.H.S = R.H.S

therefore, p(n) is true for n = 1


let us assume p(k) is true

1 + 2² + 3² + 4² + ....... k² = 1/6[k(k + 1)(2k + 1)]------( 1 )

let p(k + 1) is also true,

1 + 2² + 3² + 4² + ........ (k + 1)² = 1/6[(k + 1){(k + 1) + 1}{2(k + 1) + 1}]

1 + 2² + 3² + 4² +........ k² + (k + 1)² = 1/6[(k + 1)(k + 2)(2k + 3)]


from-------( 1 )

1/6[k(k + 1)(2k + 1)] + (k + 1)²

=> [(k + 1){(2k² + k) + 6(k + 1)}]/6

=> [(k + 1){2k² + 7k + 6}]/6

=> [(k + 1){2k² + 4k + 3k + 6}]/6

=> [(k + 1){2k(k + 2) + 3(k + 2)}]/6

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

hence, L.H.S = R.H.S


therefore, p(k + 1) = p(k) is true

thus, By principle of mathematical induction, p(n) is true for n , for all n ∈ N


I HOPE IT'S HELP YOU DEAR,
THANKS