Tag: telescopic summation for infinte series

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Questions Related to telescopic summation for infinte series

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

If y= -1  then the value of 
$\displaystyle 1+\frac{1}{y}+\frac{1}{y^{2}}+\frac{1}{y^{3}}+\frac{1}{y^{4}}+\frac{1}{y^{5}}$ is

  1. -1

  2. 0

  3. 1

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1+\frac{1}{y}+\frac{1}{y^{2}}+\frac{1}{y^{3}}+\frac{1}{y^{4}}+\frac{1}{y^{5}}$

Put the value y=-1
Then $1+\frac{1}{-1}+\frac{1}{(-1)^{2}}+\frac{1}{(-1)^{3}}+\frac{1}{(-1)^{4}}+\frac{1}{(-1)^{5}}= 1-1+1-1+1-1= 3-3=0$

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the sum of the following geometric series:
$ \sqrt{7}, \sqrt{21}, 3\sqrt{7},...$ to n terms

  1. $ \sqrt{7}\left ( \dfrac{3^{-n/2}-1}{\sqrt{3}-1} \right )$
  2. $ \sqrt{6}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
  3. $ \sqrt{7}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
  4. $ \sqrt{5}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $ S _n$ denote the sum of n terms of the G.P. $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, ...,$ 


Cleraly, the given series is a $G.P.$ with first term$=a=\sqrt7$ and common ratio$=r=\sqrt 3$

Then,
$ S _n = \sqrt{7} \left { \dfrac{\left ( \sqrt{3} \right )^{n}-1}{\sqrt{3}-1} \right } = \sqrt{7} \left ( \dfrac{3^{n/2}-1}{3^{1/2}-1} \right )$

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$ is equal to

  1. $22$
  2. $23$
  3. $44$
  4. $45$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Now,

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\sin^2 46^o+... +\sin^2 98^o)+ \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\cos^2 44^o+... +\cos^2 2^o)+ 1$ [ Since $\sin^2 x^o=\cos^2 (90^o-x^o)$]
$=(\sin^2 2^o+\cos^2 2^o)+(\sin^2 4^o+\cos^2 4^o)+......+(\sin^2 44^o+\cos^2 44^o)+1$
$=1+1+.....+1(22\text{th})+1$
$=22+1$
$=23$.

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the missing terms :- .........., .........., 8, .........., -1

  1. 16, 10 and 2

  2. 17, 12 and 3

  3. 17, 12.5 and 3.5

  4. 1, 2 and 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Third term, $a _3 = 8$
Fifth term, $a _5 = - 1$
we know that, $a _n = a + (n - 1) d$
$a _3 = a + (3 - 1) d$
$a _3 = a + 2d $     ...... (1)
$a _5 = a + (5 - 1) d$
$-1 = a + 4d $   ....... (2)
comparing equation (1) & (2)
$8    =   a    +   2d$     .... (1)
$-1   =   a    +   4d$     .... (2)
 (+)    (-)    (-)
-------------------
   $a = - 2d$
$d = - 4.5 $
Put $d = -4.5 $ in equation (1)
$8 = a + 2 (- 4.5)$
$8 = a - a$
$a = 17$
$\therefore a _2 = 17 + (2 - 1) d$
$= 17 + 1 (-4.5)$
$a _2 = 12.5$
$a _4 = 17 + (4 - 1) (-4.5)$
$a _4 = 3.5$
The missing terms are 17, 12.5 and 3.5

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

If the sum of the first n integers is 15. What is n? (use Gauss method)

  1. 7, 5

  2. 6, -5

  3. 3, -5

  4. 2, 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Gauss formula,
$S _n = \dfrac{n}{2} (n + 1)$
$15 = \dfrac{n}{2} (n + 1)$
$30 = n (n + 1)$
$30 = n^2 + n$   ....... (1)
By factorization we can write equation (1) as
$n^2 - 6n + 5n - 30 = 0$
$n(n - 6) + 5 (n - 6) = 0$
$n = 6, - 5$
$\therefore$ The n integers will start from -5 to 6.

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the sum of the first 100 terms -5, -4, -3, -2, -1, 0, 1, 2 ............. using Gauss method

  1. 4,400

  2. 4,100

  3. 4,200

  4. 4,450
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that $a = -5, n = 100, a _n =?, d = 1$
we know that Gauss formula is, $S _n = \dfrac{n}{2}$   [First term + Last term]
To find nth term,
$a _n = a + (n - 1) d$
$a _{100} = - 5 + (100 - 1)1$
$= - 5 + 99$
$a _{100} =94$
$S _n = \dfrac{100}{2} [-5 + 94]$
$= 50 [89]$
$S _{100} = 4,450$