Just (not) Binomial Theorem

Probability Level 4

( 1 + x ) 1000 + 2 x ( 1 + x ) 999 + 3 x 2 ( 1 + x ) 998 + ⋯ + 1001 x 1000 (1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}+\cdots+1001x^{1000}

If the coefficient of x 50 x^{50} in the above expression is in the form ( 1002 k ) \dbinom{1002}{k} and k < 100 k<100 , then find the positive integer k k .

Notation: ( M N ) = M ! N ! ( M − N ) ! \dbinom MN = \dfrac {M!}{N! (M-N)!} denotes the binomial coefficient .


The answer is 50.

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Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Sabhrant Sachan
Nov 4, 2016

Let S = ( 1 + x ) 1000 + 2 x ( 1 + x ) 999 + 3 x 2 ( 1 + x ) 998 ⋯ + 1001 x 1000 ( − x 1 + x ) S = − x ( 1 + x ) 999 − 2 x 2 ( 1 + x ) 998 − ⋯ − 1000 x 1000 − 1001 x 1001 1 + x Adding The two equations : 1 x + 1 S = ( 1 + x ) 1000 + x ( 1 + x ) 999 + x 2 ( 1 + x ) 998 ⋯ + x 1000 − 1001 x 1001 1 + x − x x + 1 × 1 x + 1 S = − x ( 1 + x ) 999 − x 2 ( 1 + x ) 998 − x 3 ( 1 + x ) 997 ⋯ − x 1001 1 + x + 1001 x 1002 ( 1 + x ) 2 Adding the two equations : 1 ( 1 + x ) 2 S = ( 1 + x ) 1000 − 1002 x 1001 1 + x + 1001 x 1002 ( 1 + x ) 2 S = ( 1 + x ) 1002 − 1002 x 1001 ( 1 + x ) + 1001 x 1002 co-efficient of x 50 can only be obtained from ( 1 + x ) 1002 which is ( 1002 50 ) Our answer is : k = 50 \text{Let } S= (1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}\cdots+1001x^{1000} \\ \left( \dfrac{-x}{1+x}\right)S=-x(1+x)^{999}-2x^2(1+x)^{998}-\cdots-1000x^{1000}-1001\dfrac{x^{1001}}{1+x} \\ \text{Adding The two equations : } \\ \dfrac{1}{x+1}S=(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}\cdots+x^{1000}-1001\dfrac{x^{1001}}{1+x} \\ \dfrac{-x}{x+1}\times \dfrac{1}{x+1}S =-x(1+x)^{999}-x^2(1+x)^{998}-x^3(1+x)^{997}\cdots-\dfrac{x^{1001}}{1+x}+1001\dfrac{x^{1002}}{(1+x)^2} \\ \text{Adding the two equations : } \\ \dfrac{1}{(1+x)^2}S= (1+x)^{1000}-1002\dfrac{x^{1001}}{1+x}+1001\dfrac{x^{1002}}{(1+x)^2} \\ S= (1+x)^{1002}-1002x^{1001}(1+x)+1001x^{1002} \\ \text{co-efficient of } x^{50} \text{ can only be obtained from } (1+x)^{1002} \text{ which is } \dbinom{1002}{50} \\ \text{Our answer is : } \boxed{k=50}

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Another way to do this would be to note that the required coefficient is given by the sum ∑ k = 0 50 ( k + 1 ) ( 1000 − k 50 − k ) \sum\limits_{k=0}^{50}(k+1)\dbinom{1000-k}{50-k} which can be shown to equal ( 1002 50 ) \binom{1002}{50} using Pascal's rule (to convert the sum to a telescoping series) and elementary properties of binomial coefficients.

Prasun Biswas - 4 years, 7 months ago

Awesome Solution! :)

Prakhar Bindal - 4 years, 7 months ago

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Thank you 😃 !

Sabhrant Sachan - 4 years, 7 months ago

Good solution but over rated qs a bit...(+1) tho!!!!

rajdeep brahma - 3 years ago

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