Discussion

Match each entry of List-1 with a suitable entry in List-2 and choose the correct option.

List-1		List-2
P	The sum of the series \(\sum_{n=1}^{\infty}\frac{1}{(n+2)(n+1)}\) is equal to	I	\(\frac{3}{2}\)
Q	\(\lim_{x \to 0}\left(\frac{3}{x^{2}}\int_{0}^{x}\sin(t)\,dt\right)\) is equal to	II	1
R	Let \(\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}\cos nx+b_{n}\sin nx)\) be the Fourier series expansion of the function \(f(x)=\frac{1}{2}\sin x-\frac{1}{2}\cos x+\frac{1}{\sqrt{2}}\sin 2x\), \(x \in [0, 2\pi]\). Then, \(\sum_{n=0}^{\infty}(a_{n}^{2}+b_{n}^{2})\) is equal to	III	\(\frac{1}{2}\)

Answer: C

The correct matching is P → III, Q → I, R → II.

P: the series telescopes, \(\frac{1}{(n+1)(n+2)} = \frac{1}{n+1}-\frac{1}{n+2}\), so the sum from n = 1 is \(\frac{1}{2}\) (III).
Q: \(\int_{0}^{x}\sin t\,dt = 1-\cos x \approx \frac{x^{2}}{2}\) for small x, so the limit is \(3 \times \frac{1}{2} = \frac{3}{2}\) (I).
R: the function is already a Fourier series with \(a_{1} = -\frac{1}{2}\), \(b_{1} = \frac{1}{2}\), \(b_{2} = \frac{1}{\sqrt{2}}\) and every other coefficient zero, so the sum of squares is \(\frac{1}{4}+\frac{1}{4}+\frac{1}{2} = 1\) (II).

A common slip in P is to start the telescoping at n = 0, which gives 1; the options pairing P with I or II follow from errors of that kind.

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