Bunuel
The sequence S is defined as follows for all n ≥ 1:
\(S_n=(-1)^n*\frac{1}{n(n+1)}\)
The sum of the first 10 terms of S is:
(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1
Given: \(S_n=(-1)^n*\frac{1}{n(n+1)}\)
i.e. \(S_1=(-1)^1*\frac{1}{1(1+1)} = -1/2\)
i.e. \(S_2=(-1)^2*\frac{1}{2(2+1)} = 1/6\)
i.e. \(S_3=(-1)^3*\frac{1}{3(3+1)} = -1/12\)
i.e. \(S_4=(-1)^4*\frac{1}{4(4+1)} = 1/20\)
i.e. \(S_5=(-1)^5*\frac{1}{5(5+1)} = -1/30\)
... and so on
Sum of First 10 terms = (-1/2) + (1/6) + (-1/12) + (1/20) + (-1/30) + (1/42) .... and so on
Method-1Substitute the approximate values of terms
Sum of First 10 terms = -0.5 + 0.16 - 0.083 + 0.05 - 0.033 + 0.023 - ...
Add every two terms together
Sum of First 10 terms = (-0.5 + 0.16) + (-0.083 + 0.05) + (-0.033 + 0.023) + ...
Sum of First 10 terms = (-0.34) + (-0.03) + (-0.01) + ... = approx. (-0.4) because the next terms will be very small
i.e. It falls between 0 and -0.5
Answer: Option