2015

Calculus Level 2

∑ n = 1 2015 n 2 + n + 1 ( n 2 + n ) ( n + 1 ) ! \displaystyle{\sum _{ n=1 }^{ 2015 }{ \frac { { n }^{ 2 }+n+1 }{ \left( { n }^{ 2 }+n \right) \left( n+1 \right) ! } }} can be represented in the form a b \dfrac { a }{ b } , where a a and b b are coprime positive integers. Find the value of a − b a-b .


The answer is -1.

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Joshua Yip by joshua yip , 20, Singapore

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3 solutions

Discussions for this problem are now closed

Aareyan Manzoor
Feb 27, 2015

simplify as: = n 2 + n + 1 ( n 2 + n ) ∗ ( n + 1 ) ! = ( n + 1 ) 2 n ∗ ( n + 1 ) 2 ∗ n ! − n n ∗ ( n + 1 ) ∗ ( n + 1 ) ! = 1 n ∗ n ! − 1 ( n + 1 ) ∗ ( n + 1 ) ! \begin{array}{c}=&\dfrac{n^2+n+1}{(n^2+n)*(n+1)!}\\ =&\dfrac{(n+1)^2}{n*(n+1)^2*n!}-\dfrac{n}{n*(n+1)*(n+1)!}\\ =&\dfrac{1}{n*n!}-\dfrac{1}{(n+1)*(n+1)!} \end{array} hence ∑ n = 1 2015 ( n 2 + n + 1 ( n 2 + n ) ∗ ( n + 1 ) ! ) = ∑ n = 1 2015 ( 1 n ∗ n ! ) − ∑ n = 1 2015 ( 1 ( n + 1 ) ∗ ( n + 1 ) ! ) = ∑ n = 1 2015 ( 1 n ∗ n ! ) − ∑ n = 2 2016 ( 1 n ∗ n ! ) \sum_{n=1}^{2015} (\dfrac{n^2+n+1}{(n^2+n)*(n+1)!})=\sum_{n=1}^{2015} (\dfrac{1}{n*n!})-\sum_{n=1}^{2015}(\dfrac{1}{(n+1)*(n+1)!})=\sum_{n=1}^{2015} (\dfrac{1}{n*n!})-\sum_{n=2}^{2016}(\dfrac{1}{n*n!}) we see that all terms except last and first cancels out,hence ∑ n = 1 2015 ( n 2 + n + 1 ( n 2 + n ) ∗ ( n + 1 ) ! ) = 1 1 ∗ 1 ! − 1 2016 ∗ 2016 ! = 2016 ∗ 2016 ! − 1 2016 ∗ 2016 ! \sum_{n=1}^{2015} (\dfrac{n^2+n+1}{(n^2+n)*(n+1)!})=\dfrac{1}{1*1!}-\dfrac{1}{2016*2016!}=\dfrac{2016*2016!-1}{2016*2016!} hence a − b = ( 2016 ∗ 2016 ! − 1 ) − 2016 ∗ 2016 ! = − 1 a-b=(2016*2016!-1)-2016*2016!=\boxed{-1}

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Moderator note:

Nicely done! You should be able to get the generalization:

∑ n = 1 m n 2 + n + 1 ( n 2 + n ) ( n + 1 ) ! = 1 − 1 ( m + 1 ) ( m + 1 ) ! \displaystyle{\sum _{ n=1 }^{ m }{ \frac { { n }^{ 2 }+n+1 }{ \left( { n }^{ 2 }+n \right) \left( n+1 \right) ! } }} = 1 - \frac {1}{(m+1)(m+1)!} .

Brock Brown
Feb 27, 2015

Python: 

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from fractions import Fraction as frac
from math import factorial as fac
total = 0
for n in xrange(1,2016):
    top = n*n+n+1
    bottom = (n*n+n)*fac(n+1)
    total += frac(top, bottom)
top = total.numerator
bottom = total.denominator
print "Answer:", top-bottom

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Tanishq Varshney
Feb 27, 2015

∑ n = 1 n 1 n × n ! − 1 ( n + 1 ) × ( n + 1 ) ! \displaystyle \sum_{n=1}^{n} \frac{1}{n \times n!}-\frac{1}{(n+1)\times (n+1)!}

=> 1 − 1 ( n + 1 ) ( n + 1 ) ! 1-\frac{1}{(n+1)(n+1)!}

Replace n = 2015 n=2015

we get 1 − 1 ( 2016 ) × ( 2016 ) ! 1-\frac{1}{(2016)\times (2016)!}

now compare it to a b \frac{a}{b}

b = ( 2016 ) × ( 2016 ) ! b=(2016)\times (2016)! ................................ 1 1

and the numerator is ( 2016 ) × ( 2016 ) ! − 1 (2016)\times (2016)!-1 [This is supposed to be a a ]

=> From 1 1 we get, b − 1 = a b-1=a

a − b = − 1 \boxed{a-b=-1}

Do upvote

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution has been marked wrong. You should carefully assign your variables.

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