Prove by mathematical induction that the sum of the first odd integers

1.	Prove by mathematical induction that the sum of the first odd integers is n2
Answer» $\large\underline\purple{\bold{Solution :- }}$ $\tt \: Let \: P(n) : 1 + 3 + 5 + .... + (2n - 1) = {n}^{<klux>2</klux>}$ ☆ Step :- 1 For n = 1, we get $\tt \: \longrightarrow \: 1 = {1}^{2} = 1$ $\tt\implies \:P(n) \: is \: <klux>TRUE</klux> \: for \: n \: = \: 1$ ☆ 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 $\tt \: \longrightarrow \: 1 + 3 + 5 + .... + (2k - 1)+ (2k + 1)$ $\tt \: \longrightarrow \: {k}^{2} + 2k + 1 \: \: \: \: \{using \: (i) \}$ $\tt \: \longrightarrow \: {(k + 1)}^{2}$ $\tt\implies \:P(n) \: is \: true \: for \: n \: = k + 1$ $\tt \: \therefore \: by \: the \: process \: of \: PMI \:$ $\tt\implies \: \boxed{ \purple{ \bf \: 1 + 3 + 5 + .... + (2n - 1) = {n}^{2} }}$ $\large{\boxed{\boxed{\bf{Hence, Proved}}}}$