Tag: binomial theorem, sequence and series

Questions Related to binomial theorem, sequence and series

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

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

  1. -1

  2. 0

  3. 1

  4. 2

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Correct Option: B
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 $1+\frac{1}{-1}+\frac{1}{(-1)^{2}}+\frac{1}{(-1)^{3}}+\frac{1}{(-1)^{4}}+\frac{1}{(-1)^{5}}= 1-1+1-1+1-1= 3-3=0$

Find the sum of the series: 2 + 4 + ......... + 80 using Gauss method, where n = 100

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

  1. 4,200

  2. 4,100

  3. 4,300

  4. 4,400

Show answer
Correct Option: B
Explanation:

Sum of 'n' series using Gauss method is
$S _n = \dfrac{n}{2} $  (First term + Last term)
$= \dfrac{100}{2} (2 + 80)$
$= 50 (82)$
$S _{100} = 4,100$

999, 730, 511, 344, 215, ....

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

  1. 123

  2. 126

  3. 125

  4. 130

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Correct Option: B
Explanation:

The rule is $\displaystyle 10^{3}-1$, $\displaystyle 9^{3}+1$, $\displaystyle 8^{3}-1$, $\displaystyle 7^{3}+1$, $\displaystyle 6^{3}-1$, $\displaystyle 5^{3}+1$ The number is 126

Find the Odd one among : 1, 4, 27, 16, 25, 36

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

  1. 4

  2. 27

  3. 16

  4. 25

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Correct Option: B
Explanation:

The numbers are written as $\displaystyle 1^{2}$, $\displaystyle 2^{2}$, $\displaystyle 3^{2}$, $\displaystyle 4^{2}$, $\displaystyle 5^{2}$, $\displaystyle 6^{2}$ etc It should be 9 in place of 27

Find the sum of the series: 1 + 2 + 3 + .......... + 50 using Gauss method.

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

  1. 1,275

  2. 1,200

  3. 1,100

  4. 2,000

Show answer
Correct Option: A
Explanation:

Sum of 'n' series using Gauss method is,
$S _n = \dfrac{n(n + 1)}{2}$
$\therefore$ Given n = 50
So, $\dfrac{50(50 + 1)}{2}$
$S _{50} = 1,275$

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

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

  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 )$

Show answer
Correct Option: C
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 )$

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

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

  1. $22$

  2. $23$

  3. $44$

  4. $45$

Show answer
Correct Option: B
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 $\sin^2 x^o=\cos^2 (90^o-x^o)$]
$=(\sin^2 2^o+\cos^2 2^o)+(\sin^2 4^o+\cos^2 4^o)+......+(\sin^2 44^o+\cos^2 44^o)+1$
$=1+1+.....+1(22\text{th})+1$
$=22+1$
$=23$.

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

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

  1. 16, 10 and 2

  2. 17, 12 and 3

  3. 17, 12.5 and 3.5

  4. 1, 2 and 3

Show answer
Correct Option: C
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

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

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

  1. 7, 5

  2. 6, -5

  3. 3, -5

  4. 2, 0

Show answer
Correct Option: B
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.

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

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

  1. 4,400

  2. 4,100

  3. 4,200

  4. 4,450

Show answer
Correct Option: D
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,
$a _n = a + (n - 1) d$
$a _{100} = - 5 + (100 - 1)1$
$= - 5 + 99$
$a _{100} =94$
$S _n = \dfrac{100}{2} [-5 + 94]$
$= 50 [89]$
$S _{100} = 4,450$