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Probability Level 3

2 ( 10 0 ) + 2 2 2 ( 10 1 ) + 2 3 3 ( 10 2 ) + 2 4 4 ( 10 3 ) + ⋯ + 2 11 11 ( 10 10 ) = ? 2\dbinom{10}{0}+\dfrac{2^2}{2}\dbinom{10}{1}+\dfrac{2^3}{3}\dbinom{10}{2}+\dfrac{2^4}{4}\dbinom{10}{3}+\cdots+\dfrac{2^{11}}{11}\dbinom{10}{10} =\, ?

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

Check Out the set Binomial Thm. !
3 11 − 2 11 \frac{3^{11}-2}{11} 3 11 − 1 11 \frac{3^{11}-1}{11} 2 12 11 \frac{2^{12}}{11} 2 12 − 1 11 \frac{2^{12}-1}{11} 3 11 11 \frac{3^{11}}{11} None of these choices

Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Sabhrant Sachan
May 16, 2016

Relevant wiki: JEE Binomial Theorem

we have ( 1 + x ) 10 = ( 10 0 ) + ( 10 1 ) x + ( 10 2 ) x 2 + ( 10 3 ) x 3 + ⋯ + ( 10 10 ) x 10 (1+x)^{10}=\dbinom{10}{0}+\dbinom{10}{1}x+\dbinom{10}{2}x^2+\dbinom{10}{3}x^3+\cdots+\dbinom{10}{10}x^{10}

Integrating both sides

∫ 0 x ( 1 + x ) 10 d x = ∫ 0 x ( ( 10 0 ) + ( 10 1 ) x + ( 10 2 ) x 2 + ( 10 3 ) x 3 + ⋯ + ( 10 10 ) x 10 ) d x ( 1 + x ) 11 11 − ( 1 + 0 ) 11 11 = ( 10 0 ) x + ( 10 1 ) x 2 2 + ( 10 2 ) x 3 3 + ( 10 3 ) x 4 4 + ⋯ + ( 10 10 ) x 11 11 \displaystyle\int_0^{x}(1+x)^{10}dx=\int_0^{x}(\dbinom{10}{0}+\dbinom{10}{1}x+\dbinom{10}{2}x^2+\dbinom{10}{3}x^3+\cdots+\dbinom{10}{10}x^{10})dx \\ \dfrac{(1+x)^{11}}{11}-\dfrac{(1+0)^{11}}{11}=\dbinom{10}{0}x+\dbinom{10}{1}\dfrac{x^2}{2}+\dbinom{10}{2}\dfrac{x^3}{3}+\dbinom{10}{3}\dfrac{x^4}{4}+\cdots+\dbinom{10}{10}\dfrac{x^{11}}{11}

Put x → 2 3 11 − 1 11 = 2 ( 10 0 ) + 2 2 2 ( 10 1 ) + 2 3 3 ( 10 2 ) + 2 4 4 ( 10 3 ) + ⋯ + 2 11 11 ( 10 10 ) \text{ Put }x\rightarrow 2 \\ \quad \\ \boxed{\dfrac{3^{11}-1}{11}}=2\dbinom{10}{0}+\dfrac{2^2}{2}\dbinom{10}{1}+\dfrac{2^3}{3}\dbinom{10}{2}+\dfrac{2^4}{4}\dbinom{10}{3}+\cdots+\dfrac{2^{11}}{11}\dbinom{10}{10}

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Same solution (+1) :)

Aditya Sky - 5 years ago

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Thanks Aditya 😃

Sabhrant Sachan - 5 years ago

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