An algebra problem by Cameron CHang

Algebra Level 2

2 0 + 2 1 + 2 2 + 2 3 + ⋯ + 2 9 + 2 10 = ? \Large { 2 }^{ 0 }+{ 2 }^{ 1 }+2^2+2^3+\cdots+2^9+{ 2 }^{ 10 }=?


The answer is 2047.

Cameron CHang by Cameron CHang , 19, USA

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2 solutions

Consider the expression x n − 1 x^n-1 .It can be factored as: x n − 1 = ( x − 1 ) ( x n − 1 + x n − 2 + x n − 3 + ⋯ + x 3 + x 2 + x + 1 ) x^n-1=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x^3+x^2+x+1) Setting x = 2 x=2 ,we get: 2 n − 1 = ( 2 − 1 ) ( 2 n − 1 + 2 n − 2 + 2 n − 3 + ⋯ + 2 2 + 2 + 1 ) 2 n − 1 = 2 n − 1 + 2 n − 2 + 2 n − 3 + ⋯ + 2 2 + 2 + 1 2^n-1=(2-1)(2^{n-1}+2^{n-2}+2^{n-3}+\cdots+2^2+2+1)\\ \boxed{2^n-1=2^{n-1}+2^{n-2}+2^{n-3}+\cdots+2^2+2+1} Now,simply putting n = 11 n=11 ,gives: 1 + 2 + 2 2 + 2 3 + ⋯ + 2 10 = 2 11 − 1 = 2048 − 1 = 2047 1+2+2^2+2^3+\cdots+2^{10}=2^{11}-1=2048-1=\boxed{2047}

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There is also a nifty equation that comes in handy. For a set of powers of the same number starting with power 0, and going up by 1, You can find the sum if you define n=highest number, and x=base number using the following equation. ( x n + 1 ) − 1 ) / x − 1 (x^n+1)-1)/x-1

Cameron CHang - 5 years, 7 months ago

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Same way....! :)

Siva prasad - 5 years, 6 months ago
Sam Dave
Apr 23, 2017
  • summation of terms in GP (2 to the power 11-1)=2047 denominator reduces to 2-1=1
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