List I has four entries and List II has five entries- Each entry of List I is to be correctly matched with one or more than one entries of List II- - List I List II- A Possible values of -i-i is are P -2- B If z3-z z - 0 - Q i- then possible values of z is-are- C 1-14-1-34- 8-1-3- 54- 8- 12 - R -2i- D 132-1-142-2-152-3- S 12- T 1336- -Which of the following is CORRECT combination-

(C) 1+14+1·34· 8+1·3· 54· 8· 12 +⋯⋯∞ (1+x)^n=1+nx+n(n 1)x^22!+⋯ nbsp; nbsp; Comparing both, we get nbsp; nx =14⋯ (1) ⇒ nbsp;n(n 1)x^22=332 Using equation ...

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Byju's Answer

Standard XII

Mathematics

Vn Method

List I has fo...

Question

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

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

Which of the following is CORRECT combination?

(C) \to (P); (D) \to (T)

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(C) \to (S); (D) \to (P)

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(C) \to (S); (D) \to (T)

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(C) \to (P); (D) \to (S)

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Solution

The correct option is A $(C) \to (P); (D) \to (T)$
$(C)$
$1 + \frac{1}{4} + \frac{1 \cdot 3}{4 \cdot 8} + \frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12} + \dots \dots \infty (1 + x)^{n} = 1 + n x + \frac{n (n - 1) x^{2}}{2!} + \dots$
Comparing both, we get
$n x = \frac{1}{4} \dots (1) \Rightarrow \frac{n (n - 1) x^{2}}{2} = \frac{3}{32}$
Using equation $(1)$ , we get
$\frac{n (n - 1)}{32 n^{2}} = \frac{3}{32} \Rightarrow \frac{n - 1}{n} = 3 \Rightarrow n = - \frac{1}{2} \Rightarrow x = - \frac{1}{2}$

$∴$ Required sum $= {(1 - \frac{1}{2})}^{- 1 / 2} = \sqrt{2}$
$(C) \to (P)$

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

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Similar questions

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List I

is to be correctly matched with one or more than one entries of

List II .

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List I

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Which of the following is the only CORRECT combination?

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