Et S n - -13 x -2- x - -16 x -2- x - -110 x -2- x - upto n terms- where x -0 - If lim n - S n -1- thenequals toB- tan 1C- 1D-

The correct option is C 1Here, T_n= ^ 1((n+1)(n+2)2 x+2x) ⇒ T_n=tan ^ 1(2 x(n+2)(n+1) x^2+4) ⇒ T_n=tan ^ 1((n+22) x (n+12) x1+(n+12)(n+22) x^2) ∴ T_n=tan ^ 1(( ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

et S n = -13 ...

Question

Let $S_{n} = {cot}^{- 1} (3 x + \frac{2}{x}) + {cot}^{- 1} (6 x + \frac{2}{x}) + {cot}^{- 1} (10 x + \frac{2}{x}) + \dots + upto n terms,$ where $x > 0 .$ If $lim n \to \infty S_{n} = 1,$ then $x$ equals to

\frac{π}{4}

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1

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tan 1

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cot 1

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Solution

The correct option is D $cot 1$
Here, $T_{n} = {cot}^{- 1} (\frac{(n + 1) (n + 2)}{2} x + \frac{2}{x})$
$\Rightarrow T_{n} = {tan}^{- 1} (\frac{2 x}{(n + 2) (n + 1) x^{2} + 4})$
$\Rightarrow T_{n} = {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(\frac{n + 2}{2}) x - (\frac{n + 1}{2}) x}{1 + (\frac{n + 1}{2}) (\frac{n + 2}{2}) x^{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$
$∴ T_{n} = {tan}^{- 1} ((\frac{n + 2}{2}) x) - {tan}^{- 1} ((\frac{n + 1}{2}) x)$

$S_{n} = \sum T_{n}$
So, $S_{n} = {tan}^{- 1} ((\frac{n + 2}{2}) x) - {tan}^{- 1} x$
$\Rightarrow lim n \to \infty S_{n} = \frac{π}{2} - {tan}^{- 1} x = {cot}^{- 1} x = 1$
$\Rightarrow x = cot 1$

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Similar questions

Q. If constant term in the cubic expression

n \sum k = 1 [x - \frac{1}{k}] [x - \frac{1}{k + 1}] [x - \frac{1}{k + 2}]

S_{n}

, then

lim n \to \infty S_{n}

A sequence is a function whose domain is the set of natural number. A sequence

s_{1}, s_{2} . . .

of real numbers is said to have a limit

l

{lim}_{n \to \infty} s_{n} = l

. If

l < \infty,

then

< s_{n} >

is said to convergent. The following are well known
(i)

lim n \to \infty \frac{1}{n^{p}} = 0 (p > 0)

(ii)

lim n \to \infty x^{n} = 0 (| x | < 1)

(iii) If

lim n \to \infty s_{n} = l

, then

lim n \to \infty \frac{s_{1} + s_{2} + . . . + s_{n}}{n} = l

Let

f (x) = - 1 + | x - 1 |, g (x) = 2 - | x + 1 |

then

Q. Let

U_{n} = \frac{n!}{(n + 2)!}

where

n \in N .

S_{n} = \sum_{n - 1}^{n} U_{n}

then

{lim}_{n \to \infty} S_{n}

equals :

Q. Let

S_{n} = 1 + 2 + 3 + . . . + n

and

P_{n} = \frac{S_{2}}{S_{2} - 1} \cdot \frac{S_{3}}{S_{3} - 1} \cdot \frac{S_{4}}{S_{4} - 1} \cdot \cdot \cdot \frac{S_{n}}{S_{n} - 1}

where

n \in N, (n \geq 2) .

Then

lim n \to \infty P_{n} =

Q. Assertion :If

S_{n}

denotes the sum of

n

terms a series given by

S_{n} = \frac{n (n + 1) (n + 2)}{6} \forall n \geq 1,

then

lim n \to \infty n \sum r = 1 \frac{1}{t_{r}} = 4

Reason:

t_{n} = S_{n} - S_{n - 1}

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Let Sn=cot−1(3x+2x)+cot−1(6x+2x)+cot−1(10x+2x)+⋯+upto n terms, where x>0. If limn→∞Sn=1, then x equals to

Let $S_{n} = {cot}^{- 1} (3 x + \frac{2}{x}) + {cot}^{- 1} (6 x + \frac{2}{x}) + {cot}^{- 1} (10 x + \frac{2}{x}) + \dots + upto n terms,$ where $x > 0 .$ If $lim n \to \infty S_{n} = 1,$ then $x$ equals to