Question

1³+2³+3³+...+n³=[n(n+1) /2] prove py principal of mathematical induction

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Answer 1

Appropriate Question :-

Prove by Principal of Mathematical Induction,

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

where, n is a natural number.

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

Let assume that

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

Step :- 1 For n = 1

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

$\rm :\longmapsto\: {1} = {\bigg(\dfrac{2}{2} \bigg) }^{2}$

$\rm :\longmapsto\: {1} = {\bigg(1 \bigg) }^{2}$

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

$\bf\implies \:P(1) \: is \: true$

Step :- 2 Assume that P(n) is true for n = k

[ where k is some natural number ]

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

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

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

Consider, LHS

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

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

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

[ by using Step 2 ]

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

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

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

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

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

Hence,

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

Hence,

By the Process of Principal of Mathematical Induction,

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

Answer 2

Prove by Principal of Mathematical Induction,

$\rm :\longmapsto\:{1}^{3} + {2}^{3} + {3}^{3} + - - + {n}^{3} = {\bigg(\dfrac{n(n + 1)}{2} \bigg) }^{2}:⟼1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> +−−+n </p><p>3</p><p> =( </p><p>2</p><p>n(n+1)</p><p> </p><p> ) </p><p>2</p><p> </p><p></p><p>where, n is a natural number.</p><p></p><p>\large\underline{\sf{Solution-}} </p><p>Solution−</p><p> </p><p> </p><p></p><p>Let assume that</p><p></p><p>\rm :\longmapsto\:P(n) : \: {1}^{3} + {2}^{3} + {3}^{3} + - - + {n}^{3} = {\bigg(\dfrac{n(n + 1)}{2} \bigg) }^{2}:⟼P(n):1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> +−−+n </p><p>3</p><p> =( </p><p>2</p><p>n(n+1)</p><p> </p><p> ) </p><p>2</p><p> </p><p></p><p>Step :- 1 For n = 1</p><p></p><p>\rm :\longmapsto\:P(1) : \: {1}^{3} = {\bigg(\dfrac{1(1 + 1)}{2} \bigg) }^{2}:⟼P(1):1 </p><p>3</p><p> =( </p><p>2</p><p>1(1+1)</p><p> </p><p> ) </p><p>2</p><p> </p><p></p><p>\rm :\longmapsto\: {1} = {\bigg(\dfrac{2}{2} \bigg) }^{2}:⟼1=( </p><p>2</p><p>2</p><p> </p><p> ) </p><p>2</p><p> </p><p></p><p>\rm :\longmapsto\: {1} = {\bigg(1 \bigg) }^{2}:⟼1=(1) </p><p>2</p><p> </p><p></p><p>\rm :\longmapsto\:1 = 1:⟼1=1</p><p></p><p>\bf\implies \:P(1) \: is \: true⟹P(1)istrue</p><p></p><p>Step :- 2 Assume that P(n) is true for n = k</p><p></p><p>[ where k is some natural number ]</p><p></p><p>\rm :\longmapsto\:P(k) :{1}^{3} + {2}^{3} + {3}^{3} + - - + {k}^{3} = {\bigg(\dfrac{k(k + 1)}{2} \bigg) }^{2}:⟼P(k):1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> +−−+k </p><p>3</p><p> =( </p><p>2</p><p>k(k+1)</p><p> </p><p> ) </p><p>2</p><p> </p><p></p><p>Step :- 3 We have to prove that P(n) is true for n = k + 1</p><p></p><p>\rm :\longmapsto\:{1}^{3} + {2}^{3} + {3}^{3} + - - {k}^{3} + {(k + 1)}^{3} = {\bigg(\dfrac{(k + 1)(k + 2)}{2} \bigg) }^{2}:⟼1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> +−−k </p><p>3</p><p> +(k+1) </p><p>3</p><p> =( </p><p>2</p><p>(k+1)(k+2)</p><p> </p><p> ) </p><p>2</p><p> </p><p></p><p>Consider, LHS</p><p></p><p>\rm :\longmapsto\:{1}^{3} + {2}^{3} + {3}^{3} + - - {k}^{3} + {(k + 1)}^{3}:⟼1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> +−−k </p><p>3</p><p> +(k+1) </p><p>3</p><p> </p><p></p><p>\rm \: = \: ({1}^{3} + {2}^{3} + {3}^{3} + - - {k}^{3}) + {(k + 1)}^{3}=(1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> +−−k </p><p>3</p><p> )+(k+1) </p><p>3</p><p> </p><p></p><p>\rm \: = \: {\bigg(\dfrac{k(k + 1)}{2} \bigg) }^{2} + {(k + 1)}^{3}=( </p><p>2</p><p>k(k+1)</p><p> </p><p> ) </p><p>2</p><p> +(k+1) </p><p>3</p><p> </p><p></p><p>[ by using Step 2 ]</p><p></p><p>\rm \: = \:\dfrac{ {k}^{2} {(k + 1)}^{2} }{4} + {(k + 1)}^{3}= </p><p>4</p><p>k </p><p>2</p><p> (k+1) </p><p>2</p><p> </p><p> </p><p> +(k+1) </p><p>3</p><p> </p><p></p><p>\rm \: = \: {(k + 1)}^{2}\bigg(\dfrac{ {k}^{2} }{4} + k + 1\bigg)=(k+1) </p><p>2</p><p> ( </p><p>4</p><p>k </p><p>2</p><p> </p><p> </p><p> +k+1)</p><p></p><p>\rm \: = \: {(k + 1)}^{2}\bigg(\dfrac{ {k}^{2} + 4k + 4 }{4}\bigg)=(k+1) </p><p>2</p><p> ( </p><p>4</p><p>k </p><p>2</p><p> +4k+4</p><p> </p><p> )</p><p></p><p>\rm \: = \: {(k + 1)}^{2}\bigg(\dfrac{( {k + 2)}^{2}}{4}\bigg)=(k+1) </p><p>2</p><p> ( </p><p>4</p><p>(k+2) </p><p>2</p><p> </p><p> </p><p> )</p><p></p><p>\rm \: = \:{\bigg(\dfrac{(k + 1)(k + 2)}{2} \bigg) }^{2}=( </p><p>2</p><p>(k+1)(k+2)</p><p> </p><p> ) </p><p>2</p><p> </p><p></p><p>Hence,</p><p></p><p>\bf\implies \:P(n) \: is \: true \: for \: n = k + 1⟹P(n)istrueforn=k+1</p><p></p><p>Hence,</p><p></p><p>By the Process of Principal of Mathematical Induction,</p><p></p><p>\rm :\longmapsto\:{1}^{3} + {2}^{3} + {3}^{3} + - - + {n}^{3} = {\bigg(\dfrac{n(n + 1)}{2} \bigg) }^{2}:⟼1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> +−−+n </p><p>3</p><p> =( </p><p>2</p><p>n(n+1)</p><p> </p><p> ) </p><p>2$