Coefficients

The Given equation can be re written as

$S=\sum _{ r=0 }^{ n-1 }{ { \left( -1 \right) }^{ r }\quad ^{ n }{ C }_{ r }\quad ^{ k(n-r) }{ C }_{ n } }$ for $n=4$ and $k=101$

Consider ${ \left( 1+x \right) }^{ n }=\sum _{ r=0 }^{ n }{ ^{ n }{ C }_{ r }\quad { x }^{ r } }$

replace $x$ By $-{ \left( 1+x \right) }^{ k }$ and re writing the equation as

${ \left( { \left( 1+x \right) }^{ k }-1 \right) }^{ n }=\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n-r }\left( ^{ n }{ C }_{ r } \right) { \left( 1+x \right) }^{ k(n-r) } }$

Now Comparing coefficient of ${ x }^{ n }$ in both sides $\forall$ $(n-r)>0$ The $R.H.S.$ Gives us $S$

For $L.H.S.$ differentiate it $n$ times and dividing by $n!$ then put $x=0$ and we will get $S={ \left( k \right) }^{ n }$

$=>$ $S=\sum _{ r=0 }^{ n-1 }{ { \left( -1 \right) }^{ (n-r) }\left( ^{ n }{ C }_{ r } \right) \left( ^{ k(n-r) }{ C }_{ n } \right) } ={ \left( k \right) }^{ n }$

Now in Our Case $n=4$ and $k=101$

$S=\sum _{ r=0 }^{ 3 }{ { \left( -1 \right) }^{ (4-r) }\left( ^{ 4 }{ C }_{ r } \right) \left( ^{ 101(4-r) }{ C }_{ 4 } \right) }$

${ \left( { \left( 1+x \right) }^{ 101 }-1 \right) }^{ 4 }={ \left( { \left( 1-(1+x \right) }^{ 101 } \right) }^{ 4 }=\left( ^{ 4 }{ C }_{ 0 } \right) { \left( -{ \left( 1+x \right) }^{ 101 } \right) }^{ 0 }+\left( ^{ 4 }{ C }_{ 1 } \right) { \left( -{ \left( 1+x \right) }^{ 101 } \right) }^{ 1 }+\left( ^{ 4 }{ C }_{ 2 } \right) { \left( -{ \left( 1+x \right) }^{ 101 } \right) }^{ 2 }+\left( ^{ 4 }{ C }_{ 3 } \right) { \left( -{ \left( 1+x \right) }^{ 101 } \right) }^{ 3 }+\left( ^{ 4 }{ C }_{ 4 } \right) { \left( -{ \left( 1+x \right) }^{ 101 } \right) }^{ 4 }$

Now comparing coefficients of ${ x }^{ 4 }$ in Both Sides.

$R.H.S.$ gives us S and For $L.H.S.$ we have to compute $\frac { { f }^{ (4) }\left( 0 \right) }{ 4! }$ which is ${ 4 }^{ th }$ derivative of $f\left( x \right) ={ \left( { \left( 1+x \right) }^{ 101 }-1 \right) }^{ 4 }$ divided by $4!$

${ f }^{ 1 }\left( x \right) =404{ (x+1) }^{ 100 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 3 }$

${ f }^{ 2 }\left( x \right) =40400{ (x+1) }^{ 99 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 3 }+122412{ (x+1) }^{ 200 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 2 }$

${ f }^{ 3 }\left( x \right) =3999600{ (x+1) }^{ 98 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 3 }+36723600{ (x+1) }^{ 199 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 2 }+24727224{ (x+1) }^{ 300 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 1 }$

${ f }^{ 4 }\left( x \right) =391960800{ (x+1) }^{ 97 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 3 }+8519875200{ (x+1) }^{ 198 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 2 }+14836334400{ (x+1) }^{ 299 }{ \left( { (1+x })^{ 101 }-1) \right) }^{ 1 }+2497449624{ \left( x+1 \right) }^{ 400 }$

${ f }^{ 4 }\left( 0 \right) =2497449624$

Now coefficient of ${ x }^{ 4 }$ in $f\left( x \right)$ is $\frac { { f }^{ 4 }\left( 0 \right) }{ 4! } =\frac { 2497449624 }{ 4! } =104060401={ 101 }^{ 4 }$

@aryan goyat @Rishabh Deep Singh can you tell me how you did it pl. ?

Looking for a solution