Sequences and Series Test

Result

Sequences and Series Test - 44

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Sum to n terms the following series :

A
5 + 11 + 19 + 29 + 41 + .......

B
3 + 7 + 14 + 24 + 37 +......

C
6 + 9 + 16 + 27 + 42 + .....

D
5 + 7 + 13 + 31 + 85 + ....

Solution
Question 2
1 / -0

$$\sum\limits_{r = 0}^{10} {r{.^{10}}{C_r}{{.3}^r}{{\left( { - 2} \right)}^{10 - r}}} $$ is equal

A
$$20$$

B
$$10$$

C
$$300$$

D
$$30$$

Solution
Question 3
1 / -0

The value of the expression $$\sum _{ r=0 }^{ n }{ { (-1) }^{ r } } \left( \dfrac { ^nC_r }{^{r+3}C_r } \right) $$ is

A
$$\dfrac{n(n+1)}{2}$$

B
$$\dfrac{n+3}{3}$$

C
$$\dfrac{3}{n+3}$$

D
$$\dfrac{n+2}{2}$$

Solution
Question 4
1 / -0

$$S=\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right)+\tan^{-1}\left(\dfrac{1}{n^2+3n+3}\right)+.....+\tan^{-1}\left(\dfrac{1}{1+(n+19)(n+20)}\right)$$, then $$\tan S$$ is equal to?

A
$$\dfrac{20}{401+20n}$$

B
$$\dfrac{n}{n^2+20n+1}$$

C
$$\dfrac{20}{n^2+20n+1}$$

D
$$\dfrac{n}{401+20n}$$

Solution

$$ S = \sum_{r=0}^{19} tan^{-1} \left ( \frac{1}{1+(n+r)(n+r+1)} \right ) = tan^{-1}\left ( \frac{(n+r+1)-(n+r)}{1+(n+r)(n+r+1)} \right )$$
$$ \Rightarrow S = \sum_{r=0}^{19} \left ( tan^{-1}(n+r+1)-tan^{-1}(n+r) \right )$$
$$ =tan^{-1}(n+1) - tan^{-1}(n) = tan^{-1}(n+20)-tan^{-1}(n+19) $$
$$ + tan^{-1}(n+2) - tan^{-1}(n+1)$$
$$ tan^{-1}(n+20) - tan^{-1}(n+19)$$
$$ \Rightarrow tan \, S = \frac{(n+20)-(n)}{1+n(n+20)} = \frac{20}{n^{2}+20n+1}$$
$$\Rightarrow (C)$$
Question 5
1 / -0

Evaluate:-
If $$\sum\limits_{r - 0}^n {{{\left\{ {\frac{{^n{C_{r - 1}}}}{{^n{C_r}{ + ^n}{C_{r - 1}}}}} \right\}}^3} = \frac{{25}}{{24}}} $$

A
$$3$$

B
$$6$$

C
$$4$$

D
$$5$$

Solution
Question 6
1 / -0

Sum of the series
$$S=1^{2}-2^{2}+3^{2}-4^{2}+..... -2000^{2}+2003^{2}$$ is

A
$$2007006$$

B
$$1005004$$

C
$$2000506$$

D
$$None$$

Solution
Question 7
1 / -0

If the expansion of $$\left( x+a \right) ^{ n }$$ if the sum of odd terms be P & sum of even terms be Q, prove that

A
$${ P }^{ 2 }-{ Q }^{ 2 }=({ x }^{ 2 }-{ a }^{ 2 })^{ n }$$

B
$$4PQ=(x+a)^{ 2n }-(x-a)^{ 2n }$$

C
$$P^2-Q^2=(x^2+a^2)^{n}$$

D
$$None\ of\ these$$

Solution
Question 8
1 / -0

If $$\left| x \right| <$$ 1 and $$\left| y \right| <$$ 1, then the sum of infinity of the series $$(x+y)+(x^2+xy+y^2)+(x^3+x^2y+xy^2+y^3)+.....$$ to $$\infty $$ is

A
$$\dfrac{(x+y-xy)}{(1-x)(1-y)}$$

B
$$\dfrac{(x-y+xy)}{(1-x)(1-y)}$$

C
$$\dfrac{(x+y-xy)}{(1+x)(1+y)}$$

D
$$\dfrac{x+y}{(1+x)(1+y)}$$

Solution
Question 9
1 / -0

Find the next term
$$210,209,205,196,180,?$$

A
$$138$$

B
$$77$$

C
$$155$$

D
$$327$$

Solution

$$T_1=210$$ $$,T_2=209,$$ $$T_3=205,$$ $$T_4=196,$$ $$T_5=180,$$ $$T_6=?$$

$$T_1-T_2=1^2\\$$
$$T_2-T_3=2^2\\$$
$$T_3-T_4=3^2\\$$
$$T_4-T_5=4^2\\$$
$$T_5-T_6=5^2\\$$
$$\implies T_6=T_5 - 5^2=180-25=155$$
Question 10
1 / -0

$$\sum _{ r=1 }^{ n }{ r. } $$ $$^{2n}C_r$$ is equal to

A
$$n.2^{n-1}$$

B
$$2^{2n-1}$$

C
$$2^{n-1}+1$$

D
$$n.2^{2n-1}$$

Solution