Which Is Larger 2?

Algebra Level 3

λ = 10 0 2 ⋅ 3 101 ω = ∑ n = 1 100 3 n ( 2 n − 1 ) \large \begin{aligned} \lambda & = 100^2 \cdot 3^{101} \\ \omega & = \sum_{n = 1}^{100} 3^n (2n - 1) \end{aligned}

For λ \lambda and ω \omega as defined above, relate them.

For more problems like this, try answering this set .

λ = ω \displaystyle \lambda = \omega λ > ω \displaystyle \lambda > \omega λ < ω \displaystyle \lambda < \omega

Christian Daang by Christian Daang , 20, Philippines

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1 solution

ω = ∑ n = 1 100 3 n ( 2 n − 1 ) ≤ 3 100 ∑ n = 1 100 ( 2 n − 1 ) = 3 100 ⋅ 100 2 ( 1 + 199 ) = 10 0 2 ⋅ 3 100 < 10 0 2 3 101 = λ \begin{aligned} \omega & = \sum_{n=1}^{100} 3^n(2n-1) \le 3^{100} \sum_{n=1}^{100} (2n-1) = 3^{100} \cdot \frac {100}2(1+199) = 100^2 \cdot 3^{100} < 100^2 3^{101} = \lambda \end{aligned}

⟹ λ > ω \implies \boxed{\lambda > \omega}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

nice manipulation sir. :D ~

Christian Daang - 4 years, 2 months ago

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