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# Problem on Series | SMO, 2009 | Problem No. 25

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series.

## Problem on Series | SMO Test

Given that $x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$ where each $a_r$ is an integer , $r = 0,1,2,...,n$

Find the value of n such that $a_0 +a_1 +a_2+a_3 +...........+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}$?

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### Key Concepts

Series Problem

Algebra

Challenges and Thrills - Pre - college Mathematics

## Try with Hints

If you got stuck in this sum then we can try by understanding the pattern we are using here.If we assume x = 1 thenn the expression will be

$x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$

The right hand side of this equation will be

$a_0 +a_1 + a_2 +a_3+..................+ a_n$

and the left hand side of the given equation be like

$1 + 2^2 + 2^ 3 + ......+2^n$

$1 + 2^2 + 2^ 3 + ......+2^n$ = $a_0 +a_1 + a_2 +a_3+..................+ a_n$

Try the rest of the sum..............

In the next hint we continue from the previous hint:

so the expression , $1 + 2^2 + 2^ 3 + ......+2^n$ = $2^{n+1} - 3$

Again , $a_1 = 1+2+3+..........+n = \frac {n(n+1)}{2}$

so $a_n = 1$

Now we have almost reach the final step.I am not showing it now.Try first.........

Now coming back to the last step :

$60 - \frac {n(n+1)}{2} + \frac {n(n+1}{2} + 1 = (2^{n+1} - 3$