Bunuel wrote:

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

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