Question

Use the principle of mathematical induction to prove that
${1}^{2} + { 2}^{2} + { 3}^{2} + .... + {n }^{2} = \frac{(n (n + 1)(n + 2)}{6}$
for all natural number.​

Answer 1

EXPLANATION.

Using the principle of mathematical induction.

⇒ 1² + 2² + 3² + . . . . . . . + n² = n(n + 1)(n + 2)/6.

For all natural number.

As we know that,

Let we put n = 1 in both L.H.S & R.H.S, we get.

⇒ L.H.S = 1² = 1.

⇒ R.H.S = n(n + 1)(n + 2)/6.

⇒ R.H.S = 1(1 + 1)(1 + 2)/6.

⇒ R.H.S = 1(2)(3)/6 = 6/6 = 1.

As we can see that,

Let we assume that P(k) is true for equation,

⇒ 1² + 2² + 3² + . . . . . . . + n² = n(n + 1)(n + 2)/6.

We will prove p(k + 1) is true.

⇒ 1² + 2² + 3² + . . . . . . . + (k + 1)².

⇒ k(k + 1)(k + 2)/6 + (k + 1)².

⇒ k(k + 1)(k + 2) + 6(k + 1)²/6.

⇒ k(k + 1)(k + 2) + 6(k + 1)/6.

⇒ (k + 2)(k(k + 1)) + 6(k + 1)/6.

⇒ (k + 2)(k² + k + 6k + 6)/6.

⇒ (k + 2)[k² + 6k + k + 6)]/6.

⇒ (k + 2)[k(k + 6) + 1(k + 6)]/6.

⇒ (k + 2)[(k + 1)(k + 6)]/6.

⇒ (k + 2)(k + 1)(k + 6)/6.

Hence, p(k + 1) is true when p(k) is true.

Answer 2

We can start this by,

Using the principle of mathematical induction,

Now put n = 1 in both LHS & RHS,

L.H.S = 1² = 1

R.H.S = n(n+1)(n + 2)/6

R.H.S = 1(1+1)(1+2)/6 = 1

Take that P(k) is true,

1² +2²+ 3² - - - + n² = n(n + 1)(n+2)/6

Now prove p(k + 1) is true,

→ 1² +2²+3²+ - - - + (k+1) ²

k(k+ 1)(k+ 2)/6 + (k+ 1)²

k(k+ 1)(k + 2) + 6(k + 1)/6

(k+ 2)(k²+k+6k+6)/6

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

(k+ 2)(k + 1)(k + 6)/6

So, p(k+ 1) is true, if p(k) is true.