Question

Prove the below series,

$\boxed{ \sum \limits^n_{k = 1} {k}^{3} = \dfrac{n^2(n + 1)^2}{4} }$​

Answer 1

$\large \dag$ Question :-

$\rm Prove : \sum \limits^n_{k = 1} {k}^{3} = \dfrac{n^2(n + 1)^2}{4}$

$\large \dag$ Step by step Solution :-

$\text{We\: have} \: \sum \limits^n_{k = 1} {k}^{3}$

☆ Putting Values we get ,

${ \sum \limits^n_{k = 1} {k}^{3} = {1}^{3} + {2}^{3} + {3}^{3} + \: . \: . \: . \: . + {n}^{3}}$

☆ Now we have the identity ;

$\\ \bf\red { k^2 {(k + 1)}^{2} - ( {k - 1)}^{2} = 4 {k}^{3} + k}$

Now Putting k = 1 , 2 , 3 , . . . . . , n in above identity we get,

$\\ {1}^{2} \times {2}^{2} - {0}^{2} \times {1}^{2} = 4 \times {1}^{3}$

${2}^{2} \times {3}^{2} - {1}^{2} \times {2}^{2} = 4 \times {2}^{3}$

${3}^{2} \times {4}^{2} - {2}^{2} \times {3}^{2} = 4 \times {3}^{3}$

$. \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\. \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .$

$n( {n + 1)}^{2} - (n - 1) {n}^{2} = 3 \times {n}^{2} + n$

Now Adding Columnwise we get,

$\\ \ {n}^{2} {(n + 1)}^{2} - 0 {}^{2} \times {1}^{2} = 4( {1}^{3} + {2}^{3} + {3}^{3} + ... + {n}^{3})$

$\implies n {}^{2}( n + 1 {)}^{2} = 4 \sum \limits^n_{k = 1} {k}^{3}$

$\implies\sum \limits^n_{k = 1} {k}^{3} = \frac {{n}^{2} ( {n + 1)}^{2} }{4}$

$\implies\sum \limits^n_{k = 1} {k}^{3} = \bigg( \frac{n(n + 1)}{2} \bigg)^{2}$

Therefore,

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf\sum \limits^n_{k = 1} {k}^{3} = \bigg( \frac{n(n + 1)}{2} \bigg)^{2} } }}}$

$\large \red\maltese \: \: \underline{\pink{\underline{\frak{\pmb{\text Hence\:\:Proved }}}}}$

Answer 2

$\large \dag$ Question :-

$\rm Prove : \sum \limits^n_{k = 1} {k}^{3} = \dfrac{n^2(n + 1)^2}{4}$

$\large \dag$ Step by step Solution :-

Let,

$\sum \limits^n_{k = 1} {k}^{3} = \rm S_n$

☆ Putting Values we get ,

${ \text S_n = {1}^{3} + {2}^{3} + {3}^{3} + \: . \: . \: . \: . + {n}^{3}}$

☆ Now we have the identity ;

$\\ \bf\red { k^2 {(k + 1)}^{2} - ( {k - 1)}^{2} = 4 {k}^{3} + k}$

Now Putting k = 1 , 2 , 3 , . . . . . , n in above identity we get,

$\\ {1}^{2} \times {2}^{2} - {0}^{2} \times {1}^{2} = 4 \times {1}^{3}$

${2}^{2} \times {3}^{2} - {1}^{2} \times {2}^{2} = 4 \times {2}^{3}$

${3}^{2} \times {4}^{2} - {2}^{2} \times {3}^{2} = 4 \times {3}^{3}$

$. \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\. \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .\\ . \: . \: . \: .\: . \: . \: .\: . \: . \: .\: . \: . \: .$

$n( {n + 1)}^{2} - (n - 1) {n}^{2} = 3 \times {n}^{2} + n$

Now Adding Columnwise we get,

$\\ \ {n}^{2} {(n + 1)}^{2} - 0 {}^{2} \times {1}^{2} = 4( {1}^{3} + {2}^{3} + {3}^{3} + ... + {n}^{3}$

$\implies n {}^{2}( n + 1 {)}^{2} = 4 \sum \limits^n_{k = 1} {k}^{3}$

$\implies\sum \limits^n_{k = 1} {k}^{3} = \frac {{n}^{2} ( {n + 1)}^{2} }{4}$

$\implies\sum \limits^n_{k = 1} {k}^{3} = \bigg( \frac{n(n + 1)}{2} \bigg)^{2}$

Therefore,

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf S_n = \sum \limits^n_{k = 1} {k}^{3} = \bigg( \frac{n(n + 1)}{2} \bigg)^{2} } }}}$

$\large \red\maltese \: \: \underline{\pink{\underline{\frak{\pmb{\text Hence\:\:Proved }}}}}$