Latest TCS Aptitude Question SOLUTION: What is the value of (1-(1/2)²) (1-(1/3)²) (1-(1/4)²) (1-(1/5)²)...... (1-(1/1024)²) ?

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What is the value of (1-(1/2)²) (1-(1/3)²) (1-(1/4)²) (1-(1/5)²)...... (1-(1/1024)²) ?

Asked In TCS Shreya Khandelwal (8 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

Raghul  (8 years ago)

a2-b2 = (a-b) (a+b)
The equation given after simplification becomes (1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4).... = 1/2 * 3/2 * 2/3 * 4/3 * 3/4 .... 1025/1024..
All terms except 1/2 and 1025/1024 will get cancelled.
The answer is
1/2 * 1025/1024 = 1025/2048.
Hope this helps.

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Niraj Kumar  (8 years ago)

ans=1025/2048
(a-b)^2=(a-b)(a+b)
after solving this series we get only first -ve part of first term and last +ve part of last term and remaining will be cancel
i.e 1/2*1025/1024=1025/2048
i.e

Is this solution Helpfull? Yes (1) | No (4)  

sanjeev kumar  (8 years ago)

use (a^2-b^2)=(a+b)*(a-b)
ans is 1025/2048

Is this solution Helpfull? Yes | No (1)  

venkatesh ksl  (8 years ago)

A staircase contains 12 steps. We can climb it 1 or 2 steps. But 8th step contains a snake so we cant step on it. In how many ways can we the top? Please solve this question

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Niraj Kumar  (8 years ago)

oh sorry---a^2-b^2=(a-b)(a+b)

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