Math, asked by blooberyj, 3 days ago

use mathematical induction to prove that 1^3+2^3+3^3+...+n3=(n(n+1)/2)^2 for all positive integers

Answers

Answered by mathdude500
5

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

To prove using Principal of Mathematical Induction,

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

Let assume that ,

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

Step :- 1 For n = 1

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

\rm \: P(1) :  1 =  {\bigg[\dfrac{2}{2} \bigg]}^{2}  \\

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

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

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

\rm \: 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

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

Consider LHS

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

On substituting the value from step 2, we get

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

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

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

Hence, By the Process of Principal of Mathematical Induction,

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

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