Question

By the principle of mathematical induction, prove that, for n ≥ 1
13 + 23 + 33 + ··· + n3 =
n(n + 1)
2
2

Answer 1

Appropriate Question :-

By the principle of mathematical induction, prove that

$\sf \: {1}^{3} + {2}^{3} + {3}^{3} + - - - + {n}^{3} = {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2}$

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

Let assume that

$\sf \: P(n) : {1}^{3} + {2}^{3} + {3}^{3} + - - - + {n}^{3} = {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2}$

Step : 1 For n = 1

$\rm :\longmapsto\:\sf {1}^{3} = {\bigg[\dfrac{1(1 + 1)}{2} \bigg]}^{2}$

$\rm :\longmapsto\:\sf 1 = {\bigg[\dfrac{2}{2} \bigg]}^{2}$

$\rm :\longmapsto\:\sf 1 = 1$

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

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

$\sf \: P(k) : {1}^{3} + {2}^{3} + {3}^{3} + - - - + {k}^{3} = {\bigg[\dfrac{k(k + 1)}{2} \bigg]}^{2}$

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

$\sf \: {1}^{3} + {2}^{3} + {3}^{3} + - - - + {k}^{3} + {(k + 1)}^{3} = {\bigg[\dfrac{(k + 1)(k + 1)}{2} \bigg]}^{2}$

Consider LHS

$\sf \: {1}^{3} + {2}^{3} + {3}^{3} + - - - + {k}^{3} + {(k + 1)}^{3}$

$\rm \:  =  \: {\bigg[\dfrac{k(k + 1)}{2} \bigg]}^{2} + {(k + 1)}^{3}$

$\rm \:  =  \: \dfrac{ {k}^{2} {(k + 1)}^{2} }{4} + {(k + 1)}^{3}$

$\rm \:  =  \: {(k + 1)}^{2}\bigg[\dfrac{ {k}^{2} }{4} + k + 1\bigg]$

$\rm \:  =  \: {(k + 1)}^{2}\bigg[\dfrac{ {k}^{2} + 4k + 4}{4}\bigg]$

$\rm \:  =  \: {(k + 1)}^{2}\bigg[\dfrac{ {(k + 2)}^{2}}{4}\bigg]$

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

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

Hence,

By the Process of Principal of Mathematical Induction,

$\red{\boxed{\sf \: {1}^{3} + {2}^{3} + {3}^{3} + - - - + {n}^{3} = {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2}}}$