Its your last month coming 2017!

Probability Level 4

S = ( n 1 ) − ( 1 + 1 2 ) ( n 2 ) + ( 1 + 1 2 + 1 3 ) ( n 3 ) + ⋯ + ( − 1 ) n − 1 ( 1 + 1 2 + 1 3 + ⋯ + 1 n ) ( n n ) S=\binom n1-\left(1+\frac 12\right)\binom n2+\left(1+\frac 12+\frac 13 \right)\binom n3+\cdots +(-1)^{n-1}\left(1+\frac 12+\frac 13 + \cdots +\frac 1n\right)\binom nn

Find the value of 1 S \dfrac 1S for n = 2017 n=2017 .

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


The answer is 2017.

Md Zuhair by Md Zuhair , 20, India

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1 solution

Mark Hennings
Nov 28, 2017

Since H n = ∫ 0 1 1 − x n 1 − x d x H_n \; = \; \int_0^1 \frac{1-x^n}{1-x}\,dx we see that S n = ∑ r = 1 n ( − 1 ) r − 1 ( n r ) H r = ∫ 0 1 1 1 − x ( ∑ r = 1 n ( − 1 ) r − 1 ( n r ) ( 1 − x r ) ) d x = − ∫ 0 1 1 1 − x ( ( 1 − 1 ) n − 1 − ( 1 − x ) n + 1 ) d x = ∫ 0 1 ( 1 − x ) n − 1 d x = n − 1 \begin{aligned} S_n & = \; \sum_{r=1}^n (-1)^{r-1} \binom{n}{r} H_r \; = \; \int_0^1 \frac{1}{1-x} \left(\sum_{r=1}^n (-1)^{r-1}\binom{n}{r}(1-x^r)\right)\,dx \\ & = \; -\int_0^1 \frac{1}{1-x}\big( (1-1)^n - 1 - (1-x)^n + 1\big)\,dx \;= \; \int_0^1 (1-x)^{n-1}\,dx \; = \; n^{-1} \end{aligned} making the answer S 2017 − 1 = 2017 S_{2017}^{-1} = \boxed{2017} .

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We can also use binomial inversion theorem.

Aaron Jerry Ninan - 3 years, 5 months ago

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(+1). Can you add a solution

Md Zuhair - 3 years, 5 months ago

Can u please elaborate more? @Aaron Jerry Ninan

Rahil Sehgal - 3 years, 2 months ago

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