Comb in a Sum

Calculus Level 5

Let S = ( 30 0 ) 71 − ( 30 1 ) 72 + ( 30 2 ) 73 − ⋯ + ( 30 30 ) 101 S= \dfrac{\dbinom{30}{0}}{71} - \dfrac{\dbinom{30}{1}}{72} + \dfrac{\dbinom{30}{2}}{73}- \dots +\dfrac{\dbinom{30}{30}}{101} .

What is the multiplicative inverse of the number P P , where P = S ( 100 30 ) P=S\dbinom{100}{30} ?


The answer is 101.

Satvik Golechha by Satvik Golechha , 21, India

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1 solution

Satvik Golechha
Jun 11, 2016

The binomials and denominators suggest us to let M ( x ) = ( 30 0 ) x 71 71 − ( 30 1 ) x 72 72 + ( 30 2 ) x 73 73 − ⋯ + ( 30 30 ) x 101 101 M(x)= \dfrac{\dbinom{30}{0}x^{71}}{71} - \dfrac{\dbinom{30}{1}x^{72}}{72} + \dfrac{\dbinom{30}{2}x^{73}}{73}- \dots +\dfrac{\dbinom{30}{30}x^{101}}{101} .

Thus, we seek for M ( 1 ) − M ( 0 ) M(1)-M(0) .

So, our problem reduces to ∫ 0 1 x 70 ( 1 − x ) 30 d x \int_{0}^{1} x^{70}(1-x)^{30}dx , which is easily integrable by substitutions like x = s i n 2 t x=sin^2 t , or integration by parts.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Using beta and gamma function will be make it shorter to integrate

Prakhar Bindal - 4 years, 8 months ago

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