Sequences and Series Test

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Sequences and Series Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Sum of the series $$2.3.1+3.4.4+4.5.7+...$$ up to $$n$$ terms is

A
$$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$$

B
$$\displaystyle \frac{n}{6}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$$

C
$$\displaystyle \frac{\left ( n+1 \right )}{6}\left [ 9n^{3}+46n^{2}-34 \right ]$$

D
$$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}-34 \right ]$$

Solution

$$S=2.3.1+3.4.4+4.5.7\\ S=\sum _{ r=1 }^{ n }{ \left( r+1 \right) \left( r+2 \right) \left( 3r-2 \right) } =\sum _{ r=1 }^{ n }{ \left( r+1 \right) \left( { 3r }^{ 2 }+4r-4 \right) } $$
$$\displaystyle =\sum _{ r=1 }^{ n }{ \left( { 3r }^{ 3 }+{ 7r }^{ 2 }-4 \right) } =3{ \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }+7\left( \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } \right) -4$$
$$\displaystyle =\frac { n }{ 12 } \left[ { 9n }^{ 3 }+{ 46n }^{ 2 }+51n-34 \right] $$
Question 2
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Let $$\displaystyle \left ( a_{n} \right )n\geq 1$$ be an increasing sequence of positive integers such that 1.$$\displaystyle a_{2n}=a_{n}+n$$ for all $$\displaystyle n\geq 1$$ 2.if $$\displaystyle a_{n}$$ is prime, then n is a prime. Prove that $$\displaystyle a_{n}=n,$$ for all $$\displaystyle n\geq 1.$$

A
$$\displaystyle n\geq 1.$$

B
$$\displaystyle n\ne 1.$$

C
$$\displaystyle n= 1.$$

D
None of the above

Solution

Let $$a_1=c$$. Then $$a_2=a_1+1=c+1$$,$$a_4=a_2+2=c+3$$. Since the sequence is increasing,it follows that $$a_3=c+2$$. We Prove that $$a_n=c+n1$$ for all $$ n1$$. Indeed, if $$n=2k$$ for some integer $$k$$ this follows by induction on $$k$$. Suppose that $$a_{2^k}=c+2^k1$$. Then $$a_{2^{k+1}}=a_{2.2^k}=a_{2^k}+2^k=c+2^{k+1}1$$. If $$2^k<n<2^{k+1}$$, then $$c+2^{k}1=a_{2^k}<a_{2^k+1}<...<a_n<...<a_{2^k+1}=c+2^{k+1}1$$ add this is possible only if $$a_n=c+n1$$.
Next we prove that $$c=1$$. Suppose that $$c2$$ and let $$p<q$$ be two consecutive prime numbers greater than $$c$$. We have $$a_{qc+1}=c+qc=q$$, hence $$qc+1$$ is a prime and clearly $$qc+1p$$. It follows that for any consecutive prime numbers $$p<q$$ we have $$qc+1$$. The numbers $$(c+1)!+2,(c+1)!+3...,(c+1)!+c+1$$ are all composite, hence if $$p$$ and $$q$$ are the consecutive primes such that $$p<(c+1)!+2<(c+1)!+c+1<q$$ then $$qp>c1$$, a contradiction. It follows that $$c=0$$ and $$a_n=n$$ for all $$n1$$.
Question 3
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Let $$\displaystyle V_{r}$$ denote the sum of the first r terms of an arithmetic progression (AP) whose first term is r and the common difference is $$\displaystyle (2r-1).$$ Let $$\displaystyle T_{r}=V_{r+1}-V_{r}-2$$ and $$\displaystyle Q_{r}=T_{r+1}-T_{r}$$ for $$r=1, 2, ...$$
The sum $$\displaystyle V_{1}+V_{2}+... +V_{n}$$ is

A
$$\displaystyle \frac{1}{12}n\left ( n+1 \right )\left ( 3n^{2}-n+1 \right )$$

B
$$\displaystyle \frac{1}{12}n\left ( n+1 \right )\left ( 3n^{2}+n+2 \right )$$

C
$$\displaystyle \frac{1}{2}n\left ( 2n^{2}-n+1 \right )$$

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

Solution

$$V_r = \dfrac{r}{2}(r+(r-1)(2r-1))$$

$$=\dfrac{1}{2}(2r^3 - r^2 +r)$$

$$V_1 +V_2 +V_3 +....... = \sum V_r$$

$$\sum V_r =\dfrac{1}{2}( {2 \sum r^3 - \sum r^2 + \sum r})$$

$$=\dfrac{1}{2} (2\times \dfrac{r^2 (r+1)^2}{4} - \dfrac{r (r+a)(2r+1)}{6} + \dfrac{r(r+1)}{2})$$

$$= \dfrac{1}{12} {r (r+1)(3r^2+r+2)}$$

for $$r=n$$ (i.e. sum upto n terms)

$$ = \dfrac{1}{12}{n(n+1)(3n^2 +n+2)}$$
Question 4
1 / -0

The sum of n terms of $$1^{2}\, +\, (1^{2}\, +\, 2^{2})\, +\, (1^{2}\, +\, 2^{2}\, +\, 3^{2})\, +\, ....$$.

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

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

C
$$\displaystyle \frac{1}{12}\, n(n\, +\, 1)^{2}\, (n\, +\, 2)$$

D
$$\displaystyle \frac{1}{12}n^{2}(n\, +\, 1)^{2}$$

Solution

$$S={ 1 }^{ 2 }+\left( { 1 }^{ 2 }+{ 2 }^{ 2 } \right) +\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } \right) +...$$

$$\displaystyle=\sum _{ r=1 }^{ n }{ \left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ r }^{ 2 } \right) }$$

$$\displaystyle=\sum _{ r=1 }^{ n }{ \left( \frac { r\left( r+1 \right) \left( 2r+1 \right) }{ 6 } \right) } $$

$$\displaystyle=\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ \left( r\left( { 2r }^{ 2 }+3r+1 \right) \right) } $$

$$\displaystyle=\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ \left( { 2r }^{ 3 }+3{ r }^{ 2 }+r \right) } =\frac { 2 }{ 6 } \sum _{ r=1 }^{ n }{ { r }^{ 3 } } +\frac { 3 }{ 6 } \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ { r } } $$

$$\displaystyle=\frac { 1 }{ 3 } { \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }+\frac { 1 }{ 2 } \left( \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } \right) +\frac { 1 }{ 6 } \left( \frac { n\left( n+1 \right) }{ 2 } \right) $$

$$=\displaystyle{ \left( \frac { { n }^{ 2 }\left( n+1 \right) }{ 12 } \right) }^{ 2 }+\frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 12 } +\frac { n\left( n+1 \right) }{ 12 } $$

$$=\displaystyle\frac { 1 }{ 12 } \left( n{ \left( n+1 \right) }^{ 2 }\left( n+2 \right) \right) $$
Question 5
1 / -0

$$\displaystyle 1^{2}+\left ( 1^{2}+2^{2} \right )+\left (1^{2}+2^{2}+3^{2} \right )+...$$ to $$n$$ terms.

A
$$\dfrac{1}{12}n(n+1)(2n^3+3n^2+n)$$

B
$$\dfrac{1}{12}n(n+1)(3n^3+2n^2+n)$$

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

D
$$\dfrac{1}{4}n^2(n+1)^2$$

Solution

nth term $$T_n = 1^2+2^2+3^2+.....+n^2=\sum n^2=\cfrac{1}{6}n(n+1)(2n+1)=\cfrac{1}{6}(2n^3+3n^2+n)$$

Thus required summation is $$\displaystyle =\sum T_n =\frac{1}{6}\left(\sum 2n^3+\sum 3n^2+ \sum n\right) $$

$$\displaystyle = \frac{1}{6}\left(2\frac{n^2(n+1)^2}{4}+3\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right) =\frac{1}{12}n(n+1)(2n^3+3n^2+n)$$
Question 6
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Determine the fourth degree expression in n which is equal to $$\displaystyle \sum_{r=1}^{n}r\left ( r+1 \right )\left ( 2r+3 \right ).$$

A
$$\displaystyle \frac{1}{6}\left ( 3n^{4}+16n^{3}+27n^{2}+14n \right ).$$

B
$$\displaystyle \frac{1}{6}\left ( 3n^{4}+9n^{3}+27n^{2}+36n \right ).$$

C
$$\displaystyle \frac{1}{6}\left ( 3n^{4}+36n^{3}+27n^{2}+9n \right ).$$

D
$$\displaystyle \frac{1}{6}\left ( 3n^{4}+14n^{3}+27n^{2}+16n \right ).$$

Solution

Sum $$=\displaystyle \sum_{r=1}^{n}r\left ( r+1 \right )\left ( 2r+3 \right )=\sum_{r=1}^{n}(2r^3+5r^2+3r)$$

   $$= 2\sum_{r=1}^{n}r^3+5\sum_{r=1}^{n}r^2+3\sum_{r=1}^{n}r$$

   $$= 2(\cfrac{n(n+1)}{2})^2+5\cfrac{n(n+1)(2n+1)}{6}+3\cfrac{n(n+1)}{2}$$

       $$=\displaystyle \frac{1}{6}\left ( 3n^{4}+16n^{3}+27n^{2}+14n \right ).$$
Question 7
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Observe the given multiples of 37.
$${37\times3=111}$$
$${37\times 6 =222}$$
$${37\times9=333}$$
$${37\times12=444}$$-------------------------------
Find the product of $${37\times27}$$

A
$$999$$

B
Greatest $$3$$ digit number

C
Either A or B

D
Smallest $$3$$ digit number

Solution

$${37\times3=37\times(3\times1)=111}$$
$${37\times 6=37\times(3\times2)=222}$$
$${37\times9=37\times(3\times3)=333}$$

$${37\times27=37\times(3\times9)=999}$$
Question 8
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Sum to $$20$$ terms of the series $$1.{ 3 }^{ 2 }+2.{ 5 }^{ 2 }+3.{ 7 }^{ 2 }+...$$ is

A
$$178090$$

B
$$168090$$

C
$$188090$$

D
None of these

Solution

We have,
$${ t }_{ n }=[nth$$ term of $$1,2,3,...]\times [nth$$ term of $$3,5,7,...{ ] }^{ 2 }$$

$$=n{ \left( 2n+1 \right) }^{ 2 }=4{ n }^{ 3 }+4{ n }^{ 2 }+n$$

$$\displaystyle \therefore { S }_{ n }=\sum { { t }_{ n } } =4\sum { { n }^{ 3 } } +4\sum { { n }^{ 2 } } +\sum { n } $$

$$\displaystyle =4.{ \left\{ \frac { n\left( n+1 \right) }{ 2 } \right\} }^{ 2 }+4.\frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } +\frac { n\left( n+1 \right) }{ 2 } $$

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

$$\displaystyle \therefore { S }_{ 20 }={ 20 }^{ 2 }.{ 21 }^{ 2 }+\frac { 2 }{ 3 } .20.21.41+\frac { 1 }{ 2 } .20.21=188090$$
Question 9
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Find the value of
$$\displaystyle \left ( 1-\frac{1}{2^{2}} \right )\left ( 1-\frac{1}{3^{3}} \right )\left (1 -\frac{1}{4^{2}} \right )\left ( 1-\frac{1}{5^{2}} \right )........\left ( 1-\frac{1}{9^{2}} \right )\left ( 1-\frac{1}{10^{2}} \right )$$

A
$$\displaystyle \frac{5}{12}$$

B
$$\displaystyle \frac{1}{2}$$

C
$$\displaystyle \frac{11}{20}$$

D
$$\displaystyle \frac{7}{10}$$

Solution

Given expression can be expanded as $$\displaystyle \left ( 1-\frac{1}{4} \right )\left ( 1-\frac{1}{9} \right )\left ( 1-\frac{1}{16} \right )\left ( 1-\frac{1}{25} \right )$$............$$\displaystyle \left ( 1-\frac{1}{81} \right )\left ( 1-\frac{1}{100} \right )$$
$$\displaystyle =\frac{3^{1}}{4_{1}}\times \frac{8^{2^{1}}}{9_{8_{1}}}\times \frac{15^{5^{1}}}{25_{5_{1}}}\times \frac{24^{3^{1}}}{36_{12_{1}}}\times \frac{48^{4^{1}}}{49_{7_{1}}}$$
$$\displaystyle =\frac{63^{7^{1}}}{64_{16^{1}}}\times \frac{80^{5^{1}}}{81_{9_{1}}}\times \frac{99^{11}}{100_{20}}=\frac{11}{20}$$
Question 10
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$$\displaystyle \frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + .....$$ upto n terms is

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

B
$$\displaystyle \frac{1}{3} n^{2}$$

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

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

Solution

General term of the given series is,
$$\displaystyle T_n =\frac{\sum_{k=1}^n k^2}{\sum_{k=1}^n k}=\frac{\dfrac{1}{6}n(n+1)(2n+1)}{\dfrac{1}{2}n(n+1)}=\frac{1}{3}(2n+1)$$
The required summation is,
$$\displaystyle \sum_{k=1}^nT_k=\frac{1}{3}\left(2\sum_{k=1}^n k+\sum_{k=1}^n 1 \right)=\frac{1}{3}[n(n+1)+n]=\frac{1}{3}n(n+2)$$
Hence option 'D' is correct choice.

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Sequences and Series Test - 21

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Sequences and Series Test - 21

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Question 1

$$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$$

$$\displaystyle \frac{n}{6}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$$

$$\displaystyle \frac{\left ( n+1 \right )}{6}\left [ 9n^{3}+46n^{2}-34 \right ]$$

$$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}-34 \right ]$$

Solution

Question 2

$$\displaystyle n\geq 1.$$

$$\displaystyle n\ne 1.$$

$$\displaystyle n= 1.$$

None of the above

Solution

Question 3

$$\displaystyle \frac{1}{12}n\left ( n+1 \right )\left ( 3n^{2}-n+1 \right )$$

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

$$\displaystyle \frac{1}{2}n\left ( 2n^{2}-n+1 \right )$$

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

Solution

Question 4

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

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

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

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

Solution

Question 5

$$\dfrac{1}{12}n(n+1)(2n^3+3n^2+n)$$

$$\dfrac{1}{12}n(n+1)(3n^3+2n^2+n)$$

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

$$\dfrac{1}{4}n^2(n+1)^2$$

Solution

Question 6

$$\displaystyle \frac{1}{6}\left ( 3n^{4}+16n^{3}+27n^{2}+14n \right ).$$

$$\displaystyle \frac{1}{6}\left ( 3n^{4}+9n^{3}+27n^{2}+36n \right ).$$

$$\displaystyle \frac{1}{6}\left ( 3n^{4}+36n^{3}+27n^{2}+9n \right ).$$

$$\displaystyle \frac{1}{6}\left ( 3n^{4}+14n^{3}+27n^{2}+16n \right ).$$

Solution

Question 7

$$999$$

Greatest $$3$$ digit number

Either A or B

Smallest $$3$$ digit number

Solution

Question 8

$$178090$$

$$168090$$

$$188090$$

None of these

Solution

Question 9

$$\displaystyle \frac{5}{12}$$

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{11}{20}$$

$$\displaystyle \frac{7}{10}$$

Solution

Question 10

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

$$\displaystyle \frac{1}{3} n^{2}$$

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

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

Solution

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