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Singapore Math Olympiad

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

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series. You may use sequential hints to solve the problem.

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 \(a_0 +a_1 +a_2+a_3 +………..+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}\)?

  • 2
  • 5
  • 6
  • 0

Key Concepts


Series Problem

Algebra

Check the Answer


But try the problem first…

Answer : 5

Source
Suggested Reading

Singapore Mathematics Olympiad, 2009

Challenges and Thrills – Pre – college Mathematics

Try with Hints


First hint

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…………..

Second Hint

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………

Final Step

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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