Question

Use the principle of mathematical Induction to prove the following statements for all n belongs to Natural Number.
$\bf1 + 2 + 3+ ... + n = \dfrac{1}{2}n(n + 1)$
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Answer 1

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

Let assume that

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

Step :- 1 For n = 1

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

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

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

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

Step :- 2 For n = k, Let assume that P(n) is true, where k is some natural number.

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

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

So, we have to prove that

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

Consider LHS

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

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

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

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

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

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

Hence, By the Process of Principal of Mathematical Induction,

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

Answer 2

solution

• attachment.

hope it helps