Repeated indefinite integral

Calculus Level 5

Let

f ( n ) = n times d x d x d x n times = x n a 0 + c 1 x n 1 a 1 + c 2 x n 2 a 2 + + c n a n \begin{aligned}f(n)&=\displaystyle\underbrace{\iint\cdots\int}_{n\text{ times}} \underbrace{dxdx\cdots dx}_{n\text{ times}}\\ &=\dfrac{x^n}{a_0}+\dfrac{c_1x^{n-1}}{a_1}+\dfrac{c_2x^{n-2}}{a_2}+\cdots +\dfrac{c_n}{a_n}\end{aligned}

where a i a_i and c i c_i are constants left over from performing the integration. Let g ( n ) = i = 0 n a i g(n)=\displaystyle\sum^n_{i=0}{a_i} where a i a_i are the constants in f ( n ) . f(n). Find the last three digits of g ( 10 ) . g(10).


The answer is 914.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

This is not a level 5 question. It should be marked level 4. But anyway the problem is nice.

C Anshul
Jun 30, 2018

Observe a n = n ! a_{n}=n!

Rest is easy.

i = 0 10 i ! \displaystyle\sum_{i=0}^{10} i!

(mod 1000 1000 ) gives 914 914

Advait Nene
May 27, 2018

f ( 10 ) f(10) has 10 10 indefinite integrals in a row.

This means that a 0 = 10 ! a_{0}=10! , a 1 = 9 ! a_{1}=9! , and so on, until a 10 = 0 ! a_{10}=0! .

i = 0 10 i ! 914 ( m o d 1000 ) \sum_{i=0}^{10}i! \equiv \boxed{914} \pmod{1000}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...