Question

prove by mathematical induction that the sum of the first odd integers is n2

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\tt \: Let \: P(n) : 1 + 3 + 5 + .... + (2n - 1) = {n}^{2}$

☆ Step :- 1 For n = 1, we get

$\tt \:  \longrightarrow \: 1 = {1}^{2} = 1$

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

☆ Step :- 2 For n = k, let assume that P(n) is true

$\tt \:  \longrightarrow \:1 + 3 + 5 + ... + (2k - 1) = {k}^{2} - - (i)$

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

$\tt \: 1 + 3 + 5 + ... + (2k - 1) + (2(k + 1) - 1) = {(k + 1)}^{2}$

▪︎ Consider LHS

$\tt \:  \longrightarrow \: 1 + 3 + 5 + .... + (2k - 1)+ (2k + 1)$

$\tt \:  \longrightarrow \: {k}^{2} + 2k + 1 \: \: \: \: \{using \: (i) \}$

$\tt \:  \longrightarrow \: {(k + 1)}^{2}$

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

$\tt \:  \therefore \: by \: the \: process \: of \: PMI \:$

$\tt\implies \: \boxed{ \purple{ \bf \: 1 + 3 + 5 + .... + (2n - 1) = {n}^{2} }}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$