600,000 Points Problem

Calculus Level 5

If I 1 = ∫ 0 ∞ s i n x x d x \displaystyle{I_{1}=\int^{\infty}_{0}\dfrac{sinx}{x} dx} and I 2 = ∫ 0 ∞ s i n 3 x x d x \displaystyle{I_{2}=\int^{\infty}_{0}\dfrac{sin^3x}{x} dx}

Now say k k be, k = I 1 I 2 k=\dfrac{I_{1}}{I_{2}} .

Let following summation be S S . S = ∑ n = 1 ∞ 7 n . 6 n ( 7 n − 6 n ) . ( 7 n + 1 − 6 n + 1 ) \displaystyle{S=\sum_{n=1}^{\infty}\dfrac{7^n.6^n}{(7^n-6^n).(7^{n+1}-6^{n+1})}}

If a reaction from reactant A A to product B B have 1st order forward rate constant K K and first order backward rate constant(from B back to A) K − 1 K_{-1} .

A ⇌ B \ce{A} \ce{<=>} \ce{B}

Find the concentration of B(in M)= j j after long time, if the initial concentration [ A ] o = 1 M [A]_{o}=1 M and K = 1 s e c − 1 K=1 sec^{-1} and K − 1 = 2 s e c − 1 K_{-1}=2 sec^{-1} . (Take initial concentration of B to be 0)

Input k × j × S k \times j \times S


The answer is 4.

Md Zuhair by Md Zuhair , 20, India

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2 solutions

Mark Hennings
May 20, 2019

To begin with, I 1 = ∫ 0 ∞ sin ⁡ x x d x = ∫ 0 ∞ sin ⁡ 3 x x d x = ∫ 0 ∞ 3 sin ⁡ x − 4 sin ⁡ 3 x x d x = 3 I 1 − 4 I 2 I_1 \; = \; \int_0^\infty \frac{\sin x}{x}\,dx \;=\; \int_0^\infty \frac{\sin 3x}{x}\,dx \; =\; \int_0^\infty \frac{3\sin x - 4\sin^3x}{x}\,dx \; = \; 3I_1 - 4I_2 so that I 1 = 2 I 2 I_1 = 2I_2 , and hence k = 2 k=2 .

Since 6 n × 7 n ( 7 n − 6 n ) ( 7 n + 1 − 6 n + 1 ) = 6 n 7 n − 6 n − 6 n + 1 7 n + 1 − 6 n + 1 \frac{6^n \times 7^n}{(7^n - 6^n)(7^{n+1} - 6^{n+1})} \; = \; \frac{6^n}{7^n - 6^n} - \frac{6^{n+1}}{7^{n+1} - 6^{n+1}} we deduce that ∑ n = 1 N 6 n × 7 n ( 7 n − 6 n ) ( 7 n + 1 − 6 n + 1 ) = 6 − 6 N + 1 7 N + 1 − 6 N + 1 \sum_{n=1}^N \frac{6^n \times 7^n}{(7^n - 6^n)(7^{n+1} - 6^{n+1})} \; = \; 6 - \frac{6^{N+1}}{7^{N+1} - 6^{N+1}} for any N N , and hence S = 6 S = 6 .

Finally, the differential equation − d [ A ] d t = [ A ] − 2 ( 1 − [ A ] ) = 3 [ A ] − 2 -\frac{d[A]}{dt} \; = \; [A] - 2(1-[A]) \; = \; 3[A] - 2 or d [ A ] d t + 3 [ A ] = 2 \frac{d[A]}{dt} + 3[A] \; = \; 2 with the initial condition [ A ] ( 0 ) = 1 [A](0) = 1 , has the solution [ A ] = 2 3 + 1 3 e − 3 t [A] \; = \; \tfrac23+ \tfrac13e^{-3t} and so the concentration of B B is [ B ] = 1 − [ A ] = 1 3 − 1 3 e − 3 t [B] \; = \; 1-[A] \; = \; \tfrac13 - \tfrac13e^{-3t} which converges to j = 1 3 j=\tfrac13 as t → ∞ t \to \infty .

Thus k × j × S = 2 × 6 × 1 3 = 4 k \times j \times S \,=\, 2 \times 6 \times \tfrac13 \,=\, \boxed{4} .

6 Helpful 2 Interesting 3 Brilliant 1 Confused
Iliya Hristov
Feb 29, 2020

2 Helpful 1 Interesting 1 Brilliant 0 Confused

Iliya Hristov - 7 months, 3 weeks ago

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