Sequences and Series Test

Result

Sequences and Series Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

If $$\underset{r = 1}{\overset{n}{\sum}} r (r + 1) (2r + 3) = an^4 + bn^3 + cn^2 + dn + e$$, then

A
a + c = b + d

B
e = 0

C
$$a, b - \dfrac{2}{3}, \, c - 1$$ are in A. P.

D
$$\dfrac{c}{a}$$ is an integer

Solution
Question 2
1 / -0

$$\displaystyle \sum^{n}_{r=0} (-1)^r \,{^nC_r}. \dfrac{(1 + r \ell n 10)}{(1 + \ell n 10^n)^r} =$$

A
$$0$$

B
$$\dfrac{1}{2}$$

C
$$1$$

D
$$none\ of\ these$$

Solution
Question 3
1 / -0

The sum to $$n$$ terms of the services $$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+.......$$ is -

A
$$\frac{3n}{n+1}$$

B
$$\frac{6n}{n+1}$$

C
$$\frac{9n}{n+1}$$

D
$$\frac{12n}{n+1}$$

Solution

General term $$=\dfrac{(2n+1)6}{(n)(n+1)(2n+1)}=\dfrac{6}{n(n+1)}=6\left[\dfrac{1}{n}-\dfrac{1}{n+1}\right]$$.
Sum $$=6\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}............-\dfrac{1}{n+1}\right]$$
$$=6\left[1-\dfrac{1}{n+1}\right]=\dfrac{6n}{x+1}$$
$$B$$ is correct
Question 4
1 / -0

If $$ (1^2-t_1)+ (2^2-t_2) +.......+(n^2-1)$$, then $$t_n$$ is

A
$$\frac{n}{2}$$

B
$$n -1$$

C
$$n+1$$

D
$$n$$
Question 5
1 / -0

Evaluate the definite integral:
$$\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$$

A
$$\pi /4$$

B
$$\pi /2$$

C
$$\pi /6$$

D
$$\pi /3 $$

Solution

Let $$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$$. Then

$$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}dx$$. Then,

$$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sec^{2}x}{\sec^{2}x(\sec^{2}x+3\tan^{2}x)}\ dx$$ [Diving $$Numerator$$ and $$Denominator$$ by $$\cos^{4}x$$]

$$\Rightarrow I=\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+t^{2})(1+4t^{2})}\ dt$$, where $$t=\tan x$$

$$\Rightarrow I=-\dfrac{1}{3}\displaystyle\int_{0}^{\infty}\left(\dfrac{1}{1+t^{2}}-\dfrac{4}{1+t^{2}}\right)dt=-\dfrac{1}{3}\left[\tan^{-1}t-2\tan^{-1}2t\right]_{0}^{\infty}$$

$$\Rightarrow =-\dfrac{1}{3}\left(\dfrac{\pi}{2}-\pi\right)=\dfrac{\pi}{6}$$
Question 6
1 / -0

The unit's place digit in $$(1446)^{4n + 3}$$ is

A
$$n$$

B
$$0$$

C
$$6$$

D
None of these

Solution
Question 7
1 / -0

The sum of the infinite series, $$ { 1 }^{ 2 }-\dfrac { { 2 }^{ 2 } }{ 5 } +\dfrac { { 3 }^{ 2 } }{ { 5 }^{ 2 } } -\dfrac { { 4 }^{ 2 } }{ { 5 }^{ 3 } } +\dfrac { { 5 }^{ 2 } }{ { 5 }^{ 4 } } -\dfrac { { 6 }^{ 2 } }{ { 5 }^{ 3 } } + .......... $$ is

A
$$\dfrac { 1 }{ 2 }$$

B
$$ \dfrac { 25 }{ 24 }$$

C
$$ \dfrac { 25 }{ 54 }$$

D
$$ \dfrac { 125 }{ 252 }$$

Solution
Question 8
1 / -0

Solve then inequality
$$\dfrac {x-1}{x}\geq 2$$

A
$$x\leq 1$$

B
$$x\leq -1$$

C
$$x\leq0$$

D
$$x \in R$$

Solution

Given

$$\dfrac {x-1}{x}\geq 2$$

$$x-1\geq 2x$$

$$x+1\leq 0$$

$$\implies x\leq -1$$

$$\implies x \in (-\infty ,-1]$$
Question 9
1 / -0

The last three digits in $$10!$$ are ?

A
$$800$$

B
$$700$$

C
$$500$$

D
$$600$$

Solution

The last three digits in$$ 10!$$ are,
The $$10!$$ is,
$$=10\times 9\times 8\times 7 \times 6\times 5\times 4\times 3\times 2\times 1\\$$
$$=3628800.$$

So, the last three digits of$$ 10!$$ is
$$=3628(800),$$
$$=800$$.
Question 10
1 / -0

The sum $$2 \times 5 + 5 \times 9 + 8 \times 13 + \ldots 10$$ term is

A
$$4500$$

B
$$4555$$

C
$$5454$$

D
None of these

Solution

Solution -
General term = $$(3n-1)(4n+1)$$

$$= 12n^{2}+3n-4n-1$$

$$= 12n^{2}-4n+3n-1$$

$$=12n^{2}-n-1$$

$$\displaystyle \sum_{n=1}^{10}12n^{2}-n-1$$

$$= \dfrac{12n(n+1)(2n+1)}{6}- \dfrac{n(n+1)}{2}-n$$

$$=2\times 10\times 11\times 21-55-10$$

$$= 4620-55-10$$

$$= 4555$$
B is correct