Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be correctly matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Possible value(s) of}\sqrt{i}+\sqrt{-i}\text{is (are)}& \text{(P)}& \sqrt{2}\\ \text{(B)}& \text{If}{z}^3=\overline{z}(z\ne 0),& \text{(Q)}& i\\ & \text{then possible values of}z\text{is/are}& & \\ \text{(C)}& 1+\frac{1}{4}+\frac{1\cdot 3}{4\cdot 8}+\frac{1\cdot 3\cdot 5}{4\cdot 8\cdot 12}+\cdots \cdots \infty & \text{(R)}& \sqrt{2}i\\ \text{(D)}& \frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3}+\cdots \cdots \infty & \text{(S)}& \frac{1}{2}\\ & & \text{(T)}& \frac{13}{36}\end{array}$

Which of the following is CORRECT combination?

• A. $\left(C\right)\to \left(P\right);\left(D\right)\to \left(T\right)$
• B. $\left(C\right)\to \left(S\right);\left(D\right)\to \left(P\right)$
• C. $\left(C\right)\to \left(S\right);\left(D\right)\to \left(T\right)$
• D. $\left(C\right)\to \left(P\right);\left(D\right)\to \left(S\right)$

Answer 1

The correct option is A $\left(C\right)\to \left(P\right);\left(D\right)\to \left(T\right)$

$\left(C\right)$
$1+\frac{1}{4}+\frac{1\cdot 3}{4\cdot 8}+\frac{1\cdot 3\cdot 5}{4\cdot 8\cdot 12}+\cdots \cdots \infty (1+x{)}^n=1+nx+\frac{n(n-1)x^2}{2!}+\cdots$
Comparing both, we get
$\Rightarrow nx=\frac{1}{4}\cdots \left(1\right)\Rightarrow \frac{n(n-1)x^2}{2}=\frac{3}{32}$
Using equation $\left(1\right)$, we get
$\Rightarrow \frac{n(n-1)}{32n^2}=\frac{3}{32}\Rightarrow \frac{n-1}{n}=3\Rightarrow n=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}$

$\therefore$ Required Sum
$={(1-\frac{1}{2})}^{-1/2}=\sqrt{2}$
$\left(C\right)\to \left(P\right)$

$\left(D\right)$
$\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3}+\cdots \cdots \infty$
General term of the series is
$T_n=\frac{1}{n^2+(n-2)}n=3,4,5,\dots \dots \Rightarrow T_n=\frac{1}{3}[\frac{1}{n-1}-\frac{1}{n+2}]$

Now,
$S=\sum _{n=3}^{\infty }{T}_n\Rightarrow 3S=\sum _{n=3}^{\infty }[\frac{1}{n-1}-\frac{1}{n+2}]\Rightarrow 3S=[(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{5}-\frac{1}{8})+\dots ]\Rightarrow 3S=[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}]\therefore S=\frac{13}{36}$
$\left(D\right)\to \left(T\right)$