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Algebra Level 2

If 1 + 3 + 9 + 27 + ⋯ ⏟ n terms > 1000 \underbrace{1+3+9+27+\cdots}_{n \text{ terms}} > 1000 , find the smallest value of n n .


The answer is 7.

Anurag Pandey by Anurag Pandey , 23, India

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1 solution

Let the LHS be S S . Then we have:

S = 1 + 3 + 9 + 27 + ⋯ ⏟ n terms = 3 0 + 3 1 + 3 2 + 3 3 + . . . + 3 n − 1 = ∑ k = 0 n − 1 3 k = 3 n − 1 3 − 1 = 3 n − 1 2 \begin{aligned} S & = \underbrace{1 + 3 + 9 + 27 + \cdots}_{n \text{ terms}} \\ & = 3^0 + 3^1 + 3^2 + 3^3 + ... + 3^{n-1} \\ & = \sum_{k=0}^{n-1} 3^k = \frac {3^n-1}{3-1} = \frac {3^n-1}2 \end{aligned}

Therefore,

3 n − 1 2 > 1000 3 n − 1 > 2000 3 n > 1999 n log ⁡ 3 > log ⁡ 1999 n > log ⁡ 1999 log ⁡ 3 = 6.918 ⟹ n = 7 \begin{aligned} \frac {3^n-1}2 & > 1000 \\ 3^n - 1 & > 2000 \\ 3^n & > 1999 \\ n \log 3 & > \log 1999 \\ n & > \frac {\log 1999}{\log 3} = 6.918 \\ \implies n & = \boxed{7} \end{aligned}

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