Recursion 2

Algebra Level 3

Let a 0 = 1 a_{0}=1 , a 1 = 2 a_{1}=2 and a n = 2 × a n − 1 a_{n}=2 \times a_{n-1} be an n th n^{\text{th}} term of a sequence where n > 1 n>1 and let

S = ∑ n = 0 20 a n S=\sum_{n=0}^{20} {a_{n}}

Compute S m o d 2015 S \bmod {2015} .

1551 1771 1331 1991

Zeeshan Ali by Zeeshan Ali , 25, Pakistan

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.

1 solution

Zeeshan Ali
Dec 21, 2015

The sequence of numbers generated by the given n t h n^{th} term is as under;

1 , 2 , 4 , 8 , 16 , 32 , . . . 1, 2, 4, 8, 16, 32, . . . which is basically a geometric sequence or progression with common ratio r=2. Also we have n=20.

Therefore S= a 0 ( r n + 1 − 1 ) r − 1 \frac{a_{0}(r^{n+1}-1)}{r-1} = 1 ( 2 21 − 1 ) 2 − 1 \frac{1(2^{21}-1)}{2-1} =2,097,151

and the required answer is S mod 2015 or 2,097,151 mod 2015 or 1551 \boxed{1551}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...