5 1 + 3 + 5 + . . . + ( 2 n − 1 ) = 0.00 8 − 507 \small 5^{1+3+5+...+(2n-1)} = 0.008^{-507}

Algebra Level 2

5 1 + 3 + 5 + . . . + ( 2 n − 1 ) = 0.00 8 − 507 \large 5^{1+3+5+...+(2n-1)} = 0.008^{-507}

Solve for n n in the equation above.

Hint: What can you say about the sum of consecutive odd numbers?


The answer is 39.

R. Nadhiban by R. Nadhiban , 20, Singapore

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2 solutions

Chew-Seong Cheong
Aug 20, 2017

5 1 + 3 + 5 + . . . + ( 2 n − 1 ) = 0.00 8 − 507 5 ∑ k = 1 n ( 2 k − 1 ) = ( 8 1000 ) − 507 5 2 × n ( n + 1 ) 2 − n = ( 1000 8 ) 507 5 n 2 = ( 125 ) 507 = 5 3 × 507 = 5 1521 = 5 3 9 2 ⟹ n = 39 \large \begin{aligned} 5^{1+3+5+...+(2n-1)} & = 0.008^{-507} \\ 5^{\sum_{k=1}^n(2k-1)} & = \left(\frac 8{1000}\right)^{-507} \\ 5^{2\times \frac {n(n+1)}2-n}& = \left(\frac {1000}8\right)^{507} \\ 5^{n^2} & = \left(125\right)^{507} \\ & = 5^{3\times 507} \\ & = 5^{1521} \\ & = 5^{39^2} \\ \implies n & = \boxed{39} \end{aligned}

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R. Nadhiban
Aug 18, 2017

5^1+3+5+7+9+...+(2n-1) = (0.008)^-507 1+3+5+7+...+(2n-1) = n^2 (0.008)^-507 = (( 1 125 \frac{1}{125} )^-507 =(5^-3)^-507 =5^1521

5^n^2 = 5^1521 n^2 = 1521 n = 39

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