Question

# Find the coefficient of $${ x }^{ 50 }$$ in the expression:$${ \left( 1+x \right) }^{ 1000 }+2x{ \left( 1+x \right) }^{ 999 }+3{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+....+1001{ x }^{ 1000 }$$

A
1000C50
B
1001C50
C
1002C50
D
1003C50

Solution

## The correct option is B $${ _{ }^{ 1002 }{ C } }_{ 50 }$$Let $$S= { \left( 1+x \right) }^{ 1000 }+2x{ \left( 1+x \right) }^{ 999 }+3{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+3{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+....+1001{ x }^{ 1000 }$$       .....(1)$$\displaystyle \frac { x }{ (1+x) } S= x{ \left( 1+x \right) }^{ 999 }+2{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+3{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+....+1000{ x }^{ 1000 }+\frac { 1001 }{ (1+x) } { x }^{ 1001 }$$      ....(2)On subtracting (2) from (1), we get$$\displaystyle \left( 1-\frac { x }{ 1+x } \right) S=(1+x)^{ 1000 }+x{ \left( 1+x \right) }^{ 999 }+{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+....+{ x }^{ 1000 }-\frac { 1001 }{ (1+x) } { x }^{ 1001 }$$$$\Rightarrow \displaystyle \left( \dfrac { 1 }{ 1+x } \right) S=(1+x)^{ 1000 }+x{ \left( 1+x \right) }^{ 999 }+{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+....+{ x }^{ 1000 }-\frac { 1001 }{ (1+x) } { x }^{ 1001 }$$$$\Rightarrow \displaystyle S=(1+x)^{ 1001 }+x{ \left( 1+x \right) }^{ 1000 }+{ x }^{ 2 }{ \left( 1+x \right) }^{ 999 }+{ x }^{ 3 }{ \left( 1+x \right) }^{ 998 }+....+{ x }^{ 1000 }(1+x)-1001{ x }^{ 1001 }$$$$\Rightarrow \displaystyle S=\dfrac { (1+x)^{ 1001 }\left[1-\left( \dfrac { x }{ 1+x } \right) ^{ 1001 }\right] }{ 1-\dfrac { x }{ 1+x } } -1001{ x }^{ 1001 }$$$$\Rightarrow \displaystyle S=(1+x)^{ 1002 }\left[1-\left( \frac { x }{ 1+x } \right) ^{ 1001 }\right]-1001{ x }^{ 1001 }$$$$\Rightarrow \displaystyle S=(1+x)^{ 1002 }-(1+x)x^{ 1001 }-1001{ x }^{ 1001 }$$It is clear that in above sum , only first term can have $$x^{50}$$So , coefficient of $$x^{50}$$ in $$(1+x)^{1002}$$ is $$^{1002}C_{50}$$Mathematics

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