2015 welcomes Kishlaya's Identity!

Calculus Level 5

Let P ( n ) = ∑ m = 1 ∞ ( ∏ k = 0 n 1 m + k ) P(n) = \sum_{m=1}^\infty \left(\prod_{k=0}^n \frac{1}{m+k}\right) And Q ( n ) = ∑ k = 1 n 1 P ( k ) Q(n) = \sum_{k=1}^n \frac{1}{P(k)}

Then find the last 4 4 digits of Q ( 2015 ) + 2015 Q(2015)+2015


The answer is 2014.

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Kishlaya Jaiswal by Kishlaya Jaiswal , 24, India

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4 solutions

Kishlaya Jaiswal
Feb 1, 2015

We observe Kishlaya's Identity closely and make use of the MacLaurin Expansion of log ⁡ ( 1 − x ) \log(1-x) ∑ n = 1 ∞ ∑ m = 1 ∞ x n m ( n + m m ) = − log ⁡ ( 1 − x ) \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} = -\log(1-x) ⇒ ∑ n = 1 ∞ x n ( ∑ m = 1 ∞ 1 m ( n + m m ) ) = ∑ k = 1 ∞ x k k \Rightarrow \sum_{n=1}^\infty x^n \left(\sum_{m=1}^\infty \frac{1}{m{n+m \choose m}}\right) = \sum_{k=1}^\infty \frac{x^k}{k} On comparing the coefficients, we can conclude the following ∑ m = 1 ∞ 1 m ( n + m m ) = 1 n \sum_{m=1}^\infty \frac{1}{m{n+m \choose m}} = \frac{1}{n} ⇒ ∑ m = 1 ∞ ( m − 1 ) ! ( n + m ) ! = 1 n . n ! \Rightarrow \sum_{m=1}^\infty \frac{(m-1)!}{(n+m)!} = \frac{1}{n.n!} ⇒ ∑ m = 1 ∞ ( ∏ k = 0 n 1 m + k ) = 1 n . n ! \Rightarrow \sum_{m=1}^\infty \left(\prod_{k=0}^n \frac{1}{m+k}\right) = \frac{1}{n.n!} And thus, P ( n ) = 1 n . n ! P(n) = \frac{1}{n.n!}

(I'll leave it as an exercise for the reader to compute P ( n ) P(n) by other method(s).)

Now, we compute Q ( n ) Q(n) as Q ( n ) = ∑ k = 1 n 1 P ( k ) = ∑ k = 1 n k . k ! Q(n) = \sum_{k=1}^n \frac{1}{P(k)} = \sum_{k=1}^n k.k! We can telescope the above sum as ⇒ k . k ! = ( ( k + 1 ) − 1 ) . k ! = ( k + 1 ) ! − k ! \Rightarrow k.k! = ((k+1)-1).k! = (k+1)!-k! ⇒ Q ( n ) = ∑ k = 1 n 1 P ( k ) = ∑ k = 1 n ( k + 1 ) ! − k ! = 2 ! − 1 ! + 3 ! − 2 ! + … + ( n + 1 ) ! − n ! \Rightarrow Q(n) = \sum_{k=1}^n \frac{1}{P(k)} = \sum_{k=1}^n (k+1)!-k! = 2!-1!+3!-2!+\ldots+(n+1)!-n! And thus, ⇒ Q ( n ) = ( n + 1 ) ! − 1 \Rightarrow Q(n) = (n+1)!-1 Therefore, we are just left with computing Q ( 2015 ) + 2015 ( m o d 10000 ) = 2016 ! + 2014 ( m o d 10000 ) = 2014 Q(2015)+2015 \pmod{10000} = 2016! + 2014 \pmod{10000} = \boxed{2014}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

I used Jon Haussman method to solve , enjoyed this question , we are able to see this-

P ( n ) = ∑ m = 1 ∞ ( ∏ k = 0 n 1 m + k ) P(n) = \sum_{m=1}^\infty \left(\prod_{k=0}^n \frac{1}{m+k}\right)

P ( n ) = ∑ m = 1 ∞ 1 m × ( m + 1 ) × ( m + 2 ) ⋯ ( m + n ) P(n) = \sum_{m=1}^\infty \dfrac{1}{m\times(m + 1) \times(m+2)\cdots(m+n)}

P ( n ) = ∑ m = 1 ∞ ( m − 1 ) ! ( m + n ) ! P(n) = \sum_{m=1}^\infty \dfrac{(m-1)!}{(m+n)!}

P ( n ) = 1 n . n ! ∑ m = 1 ∞ ( m − 1 ) ! n ! ( m + n − m ) ( m + n ) ! P(n) = \dfrac{1}{n.n!} \sum_{m=1}^\infty \dfrac{(m-1)!n!(m+n - m)}{(m+n)!}

P ( n ) = 1 n . n ! ∑ m = 1 ∞ ( m − 1 ) ! n ! ( m + n − 1 ) ! − m ! n ! ( m + n ) ! P(n) = \dfrac{1}{n.n!} \sum_{m=1}^\infty \dfrac{(m - 1)!n!}{(m+n-1)!} - \dfrac{m!n!}{(m+n)!}

P ( n ) = 1 n . n ! P(n) = \dfrac{1}{n.n!}

[Q(n) = n.n!]

Q ( n ) = ( n + 1 − 1 ) n ! ⟹ ( n + 1 ) ! − n ! Q(n) = (n + 1 - 1)n! \implies (n+1)! - n!

Q ( 2015 ) + 2015 = 2016 ! − 1 + 2015 = 2016 ! + 2014 Q(2015) + 2015 = 2016! - 1 + 2015 = 2016! + 2014

I think it was not necessary to take mod , we can directly say 2016! will have many zeros so the last four digits of it will be 0000 and adding 2014 we get the last four digits as 2014.

U Z - 6 years, 4 months ago

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No, by taking ( m o d 10000 ) {\pmod{10000}} , I mean exactly the same i.e. the last four digits. It doesn't makes any difference.

Kishlaya Jaiswal - 6 years, 4 months ago

FYI, I've made it such that the skill on this problem links to the corresponding wiki page. You can see this in the top right corner.

If you add problems from a wiki page, then they will be attached directly.

Calvin Lin Staff - 6 years, 4 months ago

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Oh! Thank You So Much....

Can you tell me how can I link my previous problem also?

Kishlaya Jaiswal - 6 years, 4 months ago
Akshay Bodhare
Feb 2, 2015

The above series simplifies to

1 1.2.3..... ( n + 1 ) + 1 2.3.4.5....... ( n + 2 ) . . . . . \frac {1}{1.2.3.....(n+1)}+\frac{1}{2.3.4.5.......(n+2)}.....

multiply and divide by n.

1 n ( n + 1 − 1 1.2.3.... ( n + 1 ) + n + 2 − 2 2.3.4...... ( n + 2 ) . . . ) \frac {1}{n} (\frac {n+1-1}{1.2.3....(n+1)}+\frac{n+2-2}{2.3.4......(n+2)}...)

this is a telescopic series and thus we get

[ P ( n ) = 1 n . n ! ] [P(n)=\frac {1}{n.n!}]

After that the next part is easy.

6 Helpful 0 Interesting 0 Brilliant 0 Confused

This was why I noticed 36288000 during solving task.

Lu Chee Ket - 6 years, 4 months ago
Wei Xian Lim
Feb 10, 2015

P ( n ) = ∑ m = 1 ∞ ( ∏ k = 0 n 1 m + k ) P(n)=\sum\limits_{m=1}^{\infty}\left(\prod\limits_{k=0}^{n}\frac{1}{m+k}\right)

= ∑ m = 1 ∞ 1 n ( ∏ k = 0 n − 1 1 m + k − ∏ k = 1 n 1 m + k ) \qquad\;\,=\sum\limits_{m=1}^{\infty}\frac{1}{n}\left(\prod\limits_{k=0}^{n-1}\frac{1}{m+k}-\prod\limits_{k=1}^{n}\frac{1}{m+k}\right)

= 1 n ∏ k = 0 n − 1 1 1 + k \qquad\;\,=\frac{1}{n}\prod\limits_{k=0}^{n-1}\frac{1}{1+k}

= 1 n ⋅ n ! \qquad\;\,=\frac{1}{n\cdot n!}

Q ( n ) = ∑ k = 1 n k ⋅ k ! Q(n)=\sum\limits_{k=1}^{n}k\cdot k!

= ∑ k = 1 n ( ( k + 1 ) ! − k ! ) \qquad\;\,=\sum\limits_{k=1}^{n}\left((k+1)!-k!\right)

= ( n + 1 ) ! − 1 \qquad\;\,=(n+1)!-1

Q ( 2015 ) + 2015 = 2016 ! + 2014 Q(2015)+2015=2016!+2014

Since 2016 ! 2016! has more than four trailing zeroes, the answer is 2014 \boxed{2014} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Feb 5, 2015

1/ P (n) = {1, 4, 18, 96, 600, 4320, 35280, ... , 2311256907767808000, 48658040163532800000, 1072909785605898240000, ...} Sequence can be formulated for finding all terms;

r = n + 1/ n + 2; 1/ P (n) = (n)(n!)

Q (n) = {1, 5, 23, 119, 719, 5039, 40319, ..., 121645100408831999, ...}

[Q (2015)] = [Q (21)] = 9999

9999 + 2015 = 12014

[12014] = 2014

0 Helpful 0 Interesting 0 Brilliant 0 Confused

why do u always give this sort of a solution better of nothing......(sorry)

rajat kharbanda - 6 years, 1 month ago

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