Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series.
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 \(a_0 +a_1 +a_2+a_3 +...........+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}\)?
Series Problem
Algebra
Answer : 5
Singapore Mathematics Olympiad, 2009
Challenges and Thrills - Pre - college Mathematics
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 :
\( 60 - \frac {n(n+1)}{2} + \frac {n(n+1}{2} + 1 = (2^{n+1} - 3\)
n = 5 (Answer)
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