Irregularity

Probability Level 4

( n 1 ) − ( 1 + 1 2 ) ( n 2 ) + ( 1 + 1 2 + 1 3 ) ( n 3 ) − … + ( − 1 ) n − 1 ( 1 + 1 2 + … + 1 n ) ( n n ) { n \choose 1} - \left (1 + \frac 1 2 \right ) { n \choose 2} + \left (1 + \frac 1 2 + \frac 1 3 \right ) { n \choose 3} - \ldots \\ + (-1)^{n-1} \left (1 + \frac 1 2 + \ldots + \frac 1 n \right ) { n \choose n}

If the value of the expression above equals to S S , then what can we conclude?

S = 23 S= 23 1/S is not an integer 1 S \frac{1}{S} is an integer s= 46 31 \frac{46}{31} S = 0 S=0

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Gautam Sharma by Gautam Sharma , 23, India

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1 solution

Shubham Garg
Jun 13, 2015

We can write

T r = ( − 1 ) r − 1 ( 1 + 1 2 + 1 3 + . . . . . + 1 r ) ( n r ) T_{r}=(-1)^{r-1} (1+ \frac 12+ \frac 13+.....+\frac1r ) { n \choose r}

⇒ T r = ( − 1 ) r − 1 ∫ 0 1 ( 1 + x + x 2 + x 3 . . . . . + x r − 1 ) ( n r ) \Rightarrow T_{r}=(-1)^{r-1}\int_0^1\left(1+x+x^2+x^3.....+x^{r-1}\right){n\choose r}

⇒ ∑ r = 1 n T r = ∑ r = 1 n ( − 1 ) r − 1 ∫ 0 1 ( 1 + x + x 2 + x 3 . . . . . + x r − 1 ) ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \sum_{r=1}^n(-1)^{r-1}\int_0^1\left(1+x+x^2+x^3.....+x^{r-1}\right){n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 ∑ r = 1 n ( − 1 ) r − 1 ( 1 + x + x 2 + x 3 . . . . . + x r − 1 ) ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\sum_{r=1}^n(-1)^{r-1}\left(1+x+x^2+x^3.....+x^{r-1}\right){n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 ∑ r = 1 n ( − 1 ) r − 1 ( x r − 1 x − 1 ) ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\sum_{r=1}^n(-1)^{r-1}\left(\frac{x^r-1}{x-1}\right){n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ∑ r = 1 n ( − 1 ) r − 1 ( x r − 1 ) ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}\sum_{r=1}^n(-1)^{r-1} \left(x^r-1\right){n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ∑ r = 1 n ( − 1 ) r − 1 ( x r ) ( n r ) − ( − 1 ) r − 1 ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}\sum_{r=1}^n(-1)^{r-1} \left(x^r\right){n\choose r}-(-1)^{r-1}{n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ∑ r = 1 n ( − 1 ) r − 1 ( x r ) ( n r ) + ( − 1 ) r ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}\sum_{r=1}^n(-1)^{r-1} \left(x^r\right){n\choose r}+(-1)^{r}{n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ( ∑ r = 1 n ( − 1 ) r − 1 ( x r ) ( n r ) + ∑ r = 1 n ( − 1 ) r ( n r ) ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}(\sum_{r=1}^n(-1)^{r-1} \left(x^r\right){n\choose r}+\sum_{r=1}^n(-1)^{r}{n\choose r})

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ∑ r = 1 n ( − 1 ) r − 1 ( x r ) ( n r ) + 0 \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}\sum_{r=1}^n(-1)^{r-1} \left(x^r\right){n\choose r}+0

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ∑ r = 1 n ( − 1 ) ( − 1 ) r ( x r ) ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}\sum_{r=1}^n(-1)(-1)^{r} \left(x^r\right){n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ∑ r = 1 n ( − 1 ) ( − x ) r ( n r ) \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}\sum_{r=1}^n(-1) (-x)^r{n\choose r}

⇒ ∑ r = 1 n T r = ∫ 0 1 1 x − 1 ( − 1 ) ( 1 − x ) n \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{x-1}(-1) (1-x)^n

⇒ ∑ r = 1 n T r = ∫ 0 1 1 1 − x ( 1 − x ) n \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1\frac{1}{1-x} (1-x)^n

⇒ ∑ r = 1 n T r = ∫ 0 1 ( 1 − x ) n − 1 \Rightarrow \sum_{r=1}^nT_{r}= \int_0^1 (1-x)^{n-1}

⇒ ∑ r = 1 n T r = 1 n \Rightarrow \sum_{r=1}^nT_{r}= \frac{1}{n}

⇒ S = 1 n \Rightarrow S= \frac{1}{n}

⇒ 1 S = n \Rightarrow \frac{1}{S}=n

Therefore 1 S \frac1S is an integer.

4 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice solution bro! reshare it so that others also see ur method.

Gautam Sharma - 6 years ago

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This problem recently In our AIITS-3 . Thank god i remembered the answer to this problem ! :)

Prakhar Bindal - 5 years, 4 months ago

Thank you :D

Shubham Garg - 5 years, 12 months ago

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