Sequences and Series Test

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Sequences and Series Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Find the sum of the following geometric series:
$$ 1,-a,a^2,-a^3,...$$ to n terms $$\left ( a\neq 1 \right )$$

A
$$ \dfrac{1+\left ( -a \right )^n}{1+a}$$

B
$$ \dfrac{1-\left ( a \right )^n}{1+a}$$

C
$$ \dfrac{1-\left ( -a \right )^n}{1+a}$$

D
$$ \dfrac{1+\left ( a \right )^n}{1+a}$$

Solution

Let $$S_n $$ denote the sum of n terms of the G.P. $$ 1-a, a^2, -a^3,...$$

Cleraly, the given series is a $$G.P.$$ with first term$$=a=1$$ and common ratio$$=r=-a$$

Then,
$$ S_n = 1 \left \{ \dfrac{\left ( -a \right )^n-1}{-a-1} \right \} = \dfrac{1-\left ( -1 \right )^n a^n}{1+a}$$
Question 2
1 / -0

In a sequence, $$a_{n}=n^{2}-1$$ then $$a_{n+1} $$ is equal to

A
$$a^{2}-5n$$

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

C
$$a^{2}+10n$$

D
$$n^{2}+2n$$

Solution

Given,

$$a_n=n^2-1$$

$$a_{n+1}=(n+1)^2-1$$

$$=n^2+2n+1-1$$

$$=n^2+2n$$
Question 3
1 / -0

Sum of the series
$$S=1+\dfrac{1}{2} \left ( 1+2 \right )+\dfrac{1}{3}\left ( 1+2+3 \right )+\dfrac{1}{4}\left ( 1+2+3+4 \right )+...$$ upto $$20$$ terms is

A
$$110.5$$

B
$$111$$

C
$$115$$

D
$$116.5$$

Solution

The general term for the above series is $$T_{n}=\dfrac{\sum n}{n}$$$$=\dfrac{n(n+1)}{2n}$$$$=\dfrac{n+1}{2}$$

Hence, $$S_{20}= \displaystyle \sum_{n=1} ^{n=20} T_{n}$$

$$\displaystyle \sum_{n=1} ^{n=20} \left (\dfrac{n+1}{2}\right)$$

$$\Rightarrow \left [1+\dfrac{3}{2}+\dfrac{4}{2}+.....+\dfrac{21}{2} \right ]$$

$$\Rightarrow \left [\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+.....+\dfrac{21}{2} \right ]$$

It is an A.P. with $$a=\dfrac{2}{2}, d=\dfrac{1}{2}, \ and \ n=20$$

We know that the sum of $$n$$ terms of A.P. is $$S_n=\dfrac{n}{2}\left [2a+(n-1)d \right]$$

$$\therefore \ S_{20}=\dfrac{20}{2}\left [2\times \dfrac{2}{2}+(20-1)\dfrac{1}{2} \right]$$

$$\Rightarrow S_{20}=10\left[2+\dfrac{19}{2}\right]=\dfrac{230}{2}=115$$

Hence, the answer is $$115$$
Question 4
1 / -0

The sum to infinity of $$\displaystyle\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+...$$is

A
$$\dfrac{1}{5}$$

B
$$\dfrac{7}{24}$$

C
$$\dfrac{5}{48}$$

D
$$\dfrac{3}{16}$$

Solution

$$\dfrac{1}{7}+\dfrac{2}{7^{2}}+\dfrac{1}{7^{3}}+\dfrac{2}{7^{4}}...\infty$$

$$=(\dfrac{1}{7}+\dfrac{1}{7^{3}}+\dfrac{1}{7^{5}}+...\infty)+(\dfrac{2}{7^{2}}+\dfrac{2}{7^{4}}+\dfrac{2}{7^{6}}+...\infty)$$

$$=\dfrac{\dfrac{1}{7}}{1-\dfrac{1}{7^{2}}}+2\left[\dfrac{\dfrac{1}{7^{2}}}{1-\dfrac{1}{7^{2}}}\right]$$ $$[$$Sum of infinite GP $$ =\cfrac { a }{ 1-r }]$$

$$=\dfrac{7}{48}+\dfrac{2}{48}$$

$$=\dfrac{9}{48}$$

$$=\dfrac{3}{16}$$
Question 5
1 / -0

$$3.6+6.9+9.12+...+3\mathrm{n}(3\mathrm{n}+3)=$$

A
$$\displaystyle \frac{n(n+1)(n+2)}{3}$$

B
$$3\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$$

C
$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{3}$$

D
$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+4)}{4}$$

Solution

$$\sum _{ k=1 }^{ n }{3k(3k+3)}=9\sum _{ k=1 }^{ n }{k^2} +9\sum _{ k=1 }^{ n }{k}$$

$$=9\times \dfrac{n(n+1)(2n+1)}{6}+9\times \dfrac{n(n+1)}{2}$$

$$=\dfrac{3}{2}n(n+1)(2n+1+3)$$

$$=3n(n+1)(n+2)$$
Question 6
1 / -0

$$2\cdot1^{2}+3\cdot2^{2}+4\cdot3^{2}+\dots $$ up to $$n$$ terms $$=$$

A
$$\displaystyle \frac{n(n+1)(n+2)(3n+1)}{12}$$

B
$$\displaystyle \frac{n(n+2)(n+3)(n+11)}{12}$$

C
$$\displaystyle \frac{n(n+1)(n+2)(3n-2)}{6}$$

D
$$\displaystyle \frac{n(n+2)(n+5)(n+8)}{6}$$

Solution

Let $$S$$ be the given sequence.
General term, $$t_{r}=(r+1)\cdot r^2=r^3+r^2$$
Thus, the sum would be,
$$S=\sum r^3 +\sum r^2 $$
$$\therefore S=\dfrac{n^2(n+1)^2}{4} + \dfrac{n(n+1)(2n+1)}{6}$$
Hence,
$$S=\dfrac{n(n+1)(3n^2+7n+2)}{12}$$
or
$$S=\dfrac{n(n+1)(n+2)(3n+1)}{12}$$
Question 7
1 / -0

$$1.4+2.5+...+\mathrm{n}(\mathrm{n}+3)=$$

A
$$\displaystyle \frac{n(n+3)(n+5)}{9}$$

B
$$\displaystyle \frac{n(n+1)(n+5)}{3}$$

C
$$\displaystyle \frac{n(n+5)(n+7)}{6}$$

D
$$\displaystyle \frac{n(n+3)(n+9)}{12}$$

Solution

$$\sum _{ k=1 }^{ n }{k(k+3)}=\sum _{ k=1 }^{ n }{k^2} +3\sum _{ k=1 }^{ n }{k}$$
$$= \dfrac{n(n+1)(2n+1)}{6}+3\times \dfrac{n(n+1)}{2}$$
$$=\dfrac{n(n+1)}{2}(\dfrac{2n+1}{3}+3)$$
$$=\dfrac{n(n+1)(n+5)}{3}$$
Question 8
1 / -0

$$ 1+ 3 + 6 + 10 + ...+\displaystyle \frac{(n-1)n}{2}+\frac{n(n+1)}{2}=$$

A
$$\displaystyle \frac{n(n+1)(n+2)}{3}$$

B
$$\displaystyle \frac{(n+1)(n+2)}{6}$$

C
$$\displaystyle \frac{n(n+1)(n+2)}{6}$$

D
$$\displaystyle \frac{(n+2)(n+1)^{2}}{3}$$

Solution

We can see that $${ T }_{ n }=\dfrac { n\left( n+1 \right) }{ 2 } $$
hence $$\Rightarrow { S }_{ n }=\sum _{ n=1 }^{ n }{ { T }_{ n } } \\ \Rightarrow { S }_{ n }=\sum _{ n=1 }^{ n }{ \dfrac { n\left( n+1 \right) }{ 2 } } =\dfrac { 1 }{ 2 } \left( \sum _{ n=1 }^{ n }{ { n }^{ 2 } } +\sum _{ n=1 }^{ n }{ n } \right) \\ =\dfrac { 1 }{ 2 } \left[ \dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } +\dfrac { n\left( n+1 \right) }{ 2 } \right] \\ =\dfrac { 1 }{ 2 } \left[ \dfrac { n\left( n+1 \right) \left( 2n+1+3 \right) }{ 6 } \right] \\ =\dfrac { 2 }{ 2 } \left[ \dfrac { n\left( n+1 \right) \left( n+2 \right) }{ 6 } \right] \\ =\dfrac { n\left( n+1 \right) \left( n+2 \right) }{ 6 } $$
Question 9
1 / -0

If $$S_1,S_2$$ and $$S_3$$. are the sums of first n natural numbers, their squares and their cubes respectively, then $$S_3\left (1+8S_1 \right )=$$

A
$$S_{2}^{2}$$

B
$$ 9S_2 $$

C
$$9S_{2}^{2}$$

D
None

Solution

$$ S_1\; =\; \dfrac{n(n+1)}{2} $$
$$ 1+ 8S_1 = 4n(n+1) +1 = (2n+1)^2$$
$$ S_2 = \dfrac{n(n+1)(2n+1)}{6} $$
$$\displaystyle S_3 = \left(\frac{n(n+1)}{2} \right)^2 $$

Now,
$$ S_3 \times (1+ 8 S_1) $$
$$= \dfrac{(2n+1)^2(n^2)(n+1)^2}{4} $$
$$= 9 \times \left (\dfrac{n(n+1)(2n+1)}{6} \right)^2 $$
$$ = 9 (S_2)^2$$
Hence, option C is correct.
Question 10
1 / -0

$$ 1.3+3.5+5.7+...+(2\mathrm{n}-1)(2\mathrm{n}+1)=$$

A
$$\displaystyle \frac{n(4n^{2}+6n-1)}{3}$$

B
$$\displaystyle \frac{n(3n^{2}+5n+1)}{3}$$

C
$$\displaystyle \frac{n(5n^{2}+7n-1)}{3}$$

D
$$\displaystyle \frac{n(7n^{2}-5n+1)}{3}$$

Solution

General term : $$(2k-1)(2k+1)$$

$$\sum_{k=1}^{n}(2k-1)(2k+1)$$$$=\sum_{k=1}^{n}(4k^2-1)$$

$$=\sum_{k=1}^{n}(4k^2)-\sum_{k=1}^{n}(1)$$

By using the formula,
$$=4\frac{n(n+1)(2n+1)}{6}-n$$

On simplifying, we get the desired result !

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Sequences and Series Test - 17

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Sequences and Series Test - 17

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SHARING IS CARING

Question 1

$$ \dfrac{1+\left ( -a \right )^n}{1+a}$$

$$ \dfrac{1-\left ( a \right )^n}{1+a}$$

$$ \dfrac{1-\left ( -a \right )^n}{1+a}$$

$$ \dfrac{1+\left ( a \right )^n}{1+a}$$

Solution

Question 2

$$a^{2}-5n$$

$$n^{2}-2n$$

$$a^{2}+10n$$

$$n^{2}+2n$$

Solution

Question 3

$$110.5$$

$$111$$

$$115$$

$$116.5$$

Solution

Question 4

$$\dfrac{1}{5}$$

$$\dfrac{7}{24}$$

$$\dfrac{5}{48}$$

$$\dfrac{3}{16}$$

Solution

Question 5

$$\displaystyle \frac{n(n+1)(n+2)}{3}$$

$$3\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$$

$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{3}$$

$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+4)}{4}$$

Solution

Question 6

$$\displaystyle \frac{n(n+1)(n+2)(3n+1)}{12}$$

$$\displaystyle \frac{n(n+2)(n+3)(n+11)}{12}$$

$$\displaystyle \frac{n(n+1)(n+2)(3n-2)}{6}$$

$$\displaystyle \frac{n(n+2)(n+5)(n+8)}{6}$$

Solution

Question 7

$$\displaystyle \frac{n(n+3)(n+5)}{9}$$

$$\displaystyle \frac{n(n+1)(n+5)}{3}$$

$$\displaystyle \frac{n(n+5)(n+7)}{6}$$

$$\displaystyle \frac{n(n+3)(n+9)}{12}$$

Solution

Question 8

$$\displaystyle \frac{n(n+1)(n+2)}{3}$$

$$\displaystyle \frac{(n+1)(n+2)}{6}$$

$$\displaystyle \frac{n(n+1)(n+2)}{6}$$

$$\displaystyle \frac{(n+2)(n+1)^{2}}{3}$$

Solution

Question 9

$$S_{2}^{2}$$

$$ 9S_2 $$

$$9S_{2}^{2}$$

None

Solution

Question 10

$$\displaystyle \frac{n(4n^{2}+6n-1)}{3}$$

$$\displaystyle \frac{n(3n^{2}+5n+1)}{3}$$

$$\displaystyle \frac{n(5n^{2}+7n-1)}{3}$$

$$\displaystyle \frac{n(7n^{2}-5n+1)}{3}$$

Solution

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