Visual SolutionPYQ 2024 · Jan 31 Shift 1Tricky

Question

The sum of the series

\frac{1}{1-3 \cdot 1^2+1^4} + \frac{2}{1-3 \cdot 2^2+2^4} + \frac{3}{1-3 \cdot 3^2+3^4} + \ldots

up to 10 terms is

(A)

\frac{45}{109}

(B)

-\frac{45}{109}

(C)

\frac{55}{109}

(D)

-\frac{55}{109}

Solution Path

Factor denominator, partial fractions telescope.

S = \frac{1}{2}(-1 - 1/109) = -55/109

01Question Setup

1/4

Evaluate the sum of the series involving

n^4 - 3n^2 + 1

in the denominator.

S = \;?

02Factorization

2/4

n^4 - 3n^2 + 1 = (n^2 - n - 1)(n^2 + n - 1)

. Partial fractions give a telescoping form.

n^4 - 3n^2 + 1 = (n^2 - n - 1)(n^2 + n - 1)

03Telescoping SumKEY INSIGHT

3/4

Only the first and last terms survive:

S = \frac{1}{2}\left(-1 - \frac{1}{109}\right) = -\frac{55}{109}

S = \frac{1}{2}\left(-1 - \frac{1}{109}\right) = -\frac{55}{109}

04Final Answer

4/4

The telescoping series collapses to a simple fraction.

\boxed{-\dfrac{55}{109}}

Concepts from this question1 concepts unlocked

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EASY

r^2 - 1 = (r-1)(r+1) allows cancellation with (r-1) denominators

\frac{r^2 - 1}{r - 1} = \frac{(r-1)(r+1)}{r-1} = r + 1

This algebraic identity is the final key step in many GP ratio problems. Without it, you get stuck.

GP simplificationPolynomial factoringSeries telescoping

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