Question

C
Q.10 Using the Principle of Mathematical induction Prove that for all neN
1 + 2 + 3 +
+n=1/2n(n+1)​

Answer 1

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

Let assume that,

$\rm :\longmapsto\:P(n): 1 + 2 + 3 + - - - + n = \dfrac{n(n + 1)}{2}$

Step :- 1 For n = 1

$\rm :\longmapsto\:P(n): 1 = \dfrac{1(1 + 1)}{2} = \dfrac{2}{2} = 1$

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

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

$\rm :\longmapsto\:P(k): 1 + 2 + 3 + - - - + k = \dfrac{k(k + 1)}{2}$

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

So,

$\rm :\longmapsto\:P(k + 1): 1 + 2 + 3 + - - - + (k + 1) = \dfrac{(k + 1)(k + 2)}{2}$

Consider,

$\rm :\longmapsto\:1 + 2 + 3 + - - - - - + (k + 1)$

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

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

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

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

Hence, By the Process of Principal of Mathematical Induction,

$\rm :\longmapsto\:1 + 2 + 3 + - - - + n = \dfrac{n(n + 1)}{2}$

Hence, Proved