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Let Pn = {(2^3 – 1)/(2^3 + 1)}*{(3^3 – 1)/(3^3 + 1)}*…..*{(n^3 – 1)/(n^3 + 1)}; n = 2, 3, ….. lim Pn as n - > ∞ is (a) ¾ (b) 7/11 (c) 2/3 (d) ½ Ans is c. but i dont know how to approach them.

Let Pn = {(2^3 – 1)/(2^3 + 1)}*{(3^3 – 1)/(3^3 + 1)}*…..*{(n^3 – 1)/(n^3 + 1)}; n = 2, 3, ….. lim Pn as n - > ∞ is
(a) ¾
(b) 7/11
(c) 2/3
(d) ½  
 
Ans is c. but i dont know how to approach them. 

Grade:12th pass

1 Answers

Arun
25763 Points
2 years ago
Dear student
 

Let’s factorize the numerator and denominators:

Numerator will be:

(2–1)(2^2+2+1)(3–1)(3^2+3+1)…(n-1)(n^2+n+1) = (1*7)(2*13)(3*21)…((n-1)*(n^2+n+1))

Similarly, denominator is:

(2+1)(2^2-2+1)(3+1)(3^2-3+1)…(n+1)(n^2-n+1) = (3*3)(4*7)(5*13)…((n+1)*(n^2-n+1))

Let’s look at the first element inside every parenthesis in numerator and denominator. Most of them will cancel out. Second element inside parenthesis also will be cancelling out as (k^2+k+1) = (k+1)^2 - (k+1) + 1.

So we are left with following fraction:

1*2*(n^2+n+1)/n*(n+1)*3 = (2n^2+2n+2)/(3n^2+3n)

When n tends to infinity, this will be 2/3

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