Question

By method of mathematical induction, prove that 4 + 10 + 18 …… + n(n+3) = n(n +1)(n+5) / 3

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \:P(n) : 4 + 10 + 18 + \cdots \cdots \: + n(n + 3) = \dfrac{n(n + 1)(n + 5)}{3}$

Step :- 1 For n = 1

$\rm \:P(1) : 4 = \dfrac{1(1 + 1)(1 + 5)}{3}$

$\rm \:P(1) : 4 = \dfrac{12}{3}$

$\rm \:P(1) : 4 = 4$

$\rm\implies \:P(n) \: is \: true \: for \: n = 1$

Step :- 2 Assume that P(n) is true for n = k, where k is some natural number.

$\rm \:P(k) : 4 + 10 + 18 + \cdots \cdots \: + k(k + 3) = \dfrac{k(k + 1)(k + 5)}{3}$

Step :- 3 We have to prove that P(n) is true for n = k + 1.

$\rm \:P(k + 1) : 4 + 10 + 18 + \cdots \: + k(k + 3) + (k + 1)(k + 4) = \dfrac{(k + 1)(k + 2)(k + 6)}{3}$

Consider LHS

$\rm \: 4 + 10 + 18 + \cdots \: + k(k + 3) + (k + 1)(k + 4) =$

Now, Substitute the value from Step 2, we get

$\rm \: = \: \dfrac{k(k + 1)(k + 5)}{3} + (k + 1)(k + 4)$

$\rm \: = \: \dfrac{k(k + 1)(k + 5) + 3(k + 1)(k + 4)}{3}$

$\rm \: = \: \dfrac{(k + 1)[k(k + 5) + 3(k + 4)]}{3}$

$\rm \: = \: \dfrac{(k + 1)[ {k}^{2} + 5k + 3k + 12]}{3}$

$\rm \: = \: \dfrac{(k + 1)[ {k}^{2} + 8k + 12]}{3}$

$\rm \: = \: \dfrac{(k + 1)[ {k}^{2} + 2k + 6k + 12]}{3}$

$\rm \: = \: \dfrac{(k + 1)[k(k + 2) + 6(k + 2)]}{3}$

$\rm \: = \dfrac{(k + 1)(k + 2)(k + 6)}{3}$

$\rm\implies \:P(n) \: is \: true \: for \: n = k + 1$

Hence, By the Process of Principal of Mathematical Induction,

$\bf \: 4 + 10 + 18 + \cdots \cdots \: + n(n + 3) = \dfrac{n(n + 1)(n + 5)}{3}$