Question

Using the mathematical induction technique, prove or disprove the following equations:

Answer 1

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

Let us assume that

$\rm :\longmapsto\:P(n) : \: 1 + 3 + 5 + - - - + (2n - 1) = {n}^{2}$

Step :- 1 For n = 1

$\rm :\longmapsto\:1 = {1}^{2} = 1$

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

Step :- 2

$\rm :\longmapsto\:Let \: assume \: for \: n = k, \: P(n) \: is \: true \: where \: k \in \: N$

So, we get

$\rm :\longmapsto\:P(k) : \: 1 + 3 + 5 + - - - + (2k - 1) = {k}^{2}$

Step :- 3

$\rm :\longmapsto\:Now, \: we \: have \: to \: prove\: for \: n = k + 1, \: P(n) \: is \: true$

So,

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

Consider, LHS

$\rm :\longmapsto \: 1 + 3 + 5 + - - - + (2k - 1) + (2k + 1)$

$\rm \: \: = \: \underbrace{ 1 + 3 + 5 + - - - + (2k - 1)} + (2k + 1)$

$\rm \: \: = \: \: {k}^{2} + 2k + 1$

$\: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:using \: step \: 2 \bigg \}}$

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

= RHS

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

So,

By the Process of Principal of Mathematical Induction,

$\: \: \: \: \: \: \underbrace{\boxed{ \bf{ \: 1 + 3 + 5 + - - - + (2n - 1) = {n}^{2} }}}$

Answer 2

Step-by-step explanation:

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