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CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 12 - MCQExams.com

 
1060640_fb38d500c9a944939a968f51305f26e1.png
  • 6
  • 7
  • 8
  • 9
  • 10
Find sum of series:
1.3.5+3.5.7+5.7.9..... ?
  • Sn=(2n1)(2n+1)(2n+3)(2n+5)8+158
  • Sn=(2n1)(2n+1)(2n+3)(2n+5)18+1518
  • Sn=(2n1)(2n+1)(2n+3)(2n+5)20+158
  • Sn=(2n1)(2n+1)(2n+3)(2n+5)6+18
sin(12)+sin(216)++sin(nn1n(n+1))+=
  • π
  • π2
  • π4
  • 3π2
The sum of the series 1+11!.14+132!(14)2+1353!(14)3+........ to is ?
  • 2
  • 2
  • 12
  • 3
The value of (21C110C1)+(21C210C2)+(3C110C3)+(21C410C4)+.....(21C1010C10) is
  • 220210
  • 221211
  • 221210
  • 22029
If an=nr=01nCr, the value of nr=0n2rnCr
  • n2an
  • 12an
  • nan
  • 0
The value of 100r=23r(22r)(r+1)(r+2) is equal to?
  • 123100100(101)
  • 323101101(102)
  • 323100100(101)
  • None of these
The sum of n terms of the series 13+33+53+ is:
  • 3n2(2n21)
  • n2(2n21)
  • n2(n21)
  • n2(2n25)
If f(x)=Π3i=1(xai)+3i=1ai3x where ai<ai+1 for i=1,2, then f(x)=0 has:
  • Only one distinct real root
  • Exactly two distinct real root
  • Exactly 3 distinct real roots
  • 3 equal real roots
The sum of 72×3(13)+93×4(13)2+114×5(13)3+ upto 10 terms is equal  to
  • 12112×310
  • 13112×310
  • 12110×310
  • None of these
The sum of the series 
(2)2+2(4)2+3(6)2+.... upto 10 terms is
  • 12100
  • 11300
  • 11200
  • 12300
The value of 10x=1r=x1r=0(2x2r) is
  • 16392
  • 16398
  • 15462
  • 15468
Observe the pattern carefully
11×11=121
111×111=12321
1111×1111=?
  • 12345321
  • 123421
  • 1234321
  • 12468
Sum of the series 
(1×2015)+(2×2014)+(3×2013)......+(2015×1) is equal to-
  • 336 ×2015×2016
  • 336 ×2015×2017
  • 336 ×2016×201
  • None
It is known  that r=11(2r1)2=π28, then r=11r2 is equal to 
  • π224
  • π23
  • π26
  • none of these
Find the sum of the infinite series 19+181+1729+....
  • 18
  • 14
  • 15
  • 23
The sum  31.2,31.2,12,42.3(12)2+53.4(12)2
  • 11(n+1)2n
  • 11n.2n1
  • 1=1(n+1)2n
  • 1(n1)2n1
If f(x)= x+12x+112x+...
then the value of f (2011). f'(2011) is :
  • 0
  • 1
  • 2011
  • 2010
1+n2+n(n1)2.4+n(n1)(n2)2.4.6+= ?
  • (23)n
  • (32)n
  • (34)n
  • (43)n
Find the missing number, if same rule is followed in all the three figures.
1180459_2279085d5b404909b0098a54eab4e625.JPG
  • 12
  • 16
  • 14
  • 9
Choose the CORRECT options:-

  • 360k=1(1kk+1+(k+1)k) is the ratio of two relative prime positive integers m nad n. The value of (m+n) is equal to 37.
  • If 5π2<x<3π, then the value of the expression 1sinx+1+sinx1sinx1+sinx is tanx2.
  • The exact value of 96sin80sin65sin35sin20+sin50+sin110 is equal to 24.
  • The sum n=1(nn4+4) is equal to 3/16.
nr=1sin1(rr1r(r+1)) is equal to 
  • tan1(n)π4
  • tan1(n+1)π4
  • tan1(n)
  • tan1(n+1)
Direction: In a given question, one number is missing in the series. you have to understand the pattern of the series and insert the number 
JN    28   27    GP
CE   12   45    TU
LR     ?     ?     MS
  • 34,36
  • 35,35
  • 30,32
  • 30,41
Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 1 + 2.22 + 32 + 2.42 + 52+2.62 +......... Find the value of A.
  • 496
  • 232
  • 248
  • 464
If 12+22+32++(2003)2=(2003)(4007)(334) and (1)(2003)+(2)(2002)=(2003)(334)(x) then x equals
  • 2005
  • 2004
  • 2003
  • 2001
The number of the three digits numbers having only two consecutive digits identical is:
  • 153
  • 160
  • 180
  • 161
The sum of the first n terms of the series 12+34+78+1516+ is equal to:
  • 2nn1
  • 12n
  • n+2n1
  • 2n1
1+(1+2)+(1+2+22)+(1+2+22+23)=.... upto n terms =______
  • 2n+2n4
  • 2(2n1)n
  • 2n+1n
  • 22n+11
If 24+37=7,12+18=3, then 54+21=?
  • 11
  • 9
  • 12
  • 3
In the following number series,one number is wrong.find out the ?

1,2,8,33,149,765,4626

  • 33
  • 8
  • 149
  • 0
If  denotes ÷, ÷ denotes ×, + denotes' - 'and × denotes +, then find the value of 116+9÷524×5.
  • 16
  • 8
  • 9
  • 4
1+12(1+2)+13(1+2+3)+14(1+2+3+4)+..... upto 20 terms is
  • 110
  • 111
  • 115
  • 116
Find the missing term in the series given below. 12, 13, 18, 19, 24, 25?
  • 27
  • 30
  • 29
  • 26
Let a sequence whose nth term is an be define as a1=12 and (n1)an1=(n+1)an for n2 ; then 
  • an=1n(n+1)
  • Sn=11n+1
  • limnSn=1
  • Sn=1n+1
The sum of the series 1.2.3+2.3.4+...3.4.5.+...to n terms is 
  • n(n+1)(n+2)
  • (n+1)(n+2)(n+3)
  • \frac { 1 }{ 4 } n(n+1)(n+2)(n+3)
  • \frac { 1 }{ 4 } (n+1)(n+2)(n+3)
The sum of first 9 terms of the series
\dfrac { { 1 }^{ 3 } }{ 1 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 } }{ 1+3 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 } }{ 1+3+5 } +\dots
  • 96
  • 142
  • 192
  • 71
If \left( { 1-y } \right) ^{ 35 }\left( 1+y \right) ^{ 45 }={ A }_{ 0 }+{ A }_{ 1 }Y+{ A }_{ 2 }{ Y }^{ 2 }+{ A }_{ 3 }{ Y }^{ 3 }+.....+{ A }_{ 80 }{ Y }^{ 80 }, then
  • \frac { { A }_{ 2 } }{ A_{ 1 } } <2
  • { A }_{ 1 }={ A }_{ 2 }
  • \frac { { A }_{ 2 } }{ { A }_{ 1 } } <1
  • 1<\frac { { A }_{ 2 } }{ { A }_{ 1 } } <2
If \dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+........\infty =\dfrac{\pi^{2}}{6} then \dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+......\infty
  • \dfrac{\pi^{2}}{8}
  • \dfrac{\pi^{2}}{12}
  • \dfrac{\pi^{2}}{3}
  • \dfrac{\pi^{2}}{2}
Sum of the series 
1+2.2+3.2^{2}+...+100.2^{10}=
  • 100.2^{100}+1
  • 99.2^{100}+1
  • 99.2^{100}-1
  • 100.2^{100}-1
Sum of series \displaystyle \sum^{n}_{r=1}(r^{2}+1)_{r!} is
  • (n+1)!
  • (n+2)!-1
  • n(n+2)!
  • (n+1)!-1
\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+...+\dfrac {1}{(3n-5)(3n-2)}
  • \dfrac {n-1}{3n+2}
  • \dfrac {n-1}{3n-2}
  • \dfrac {n-1}{3n-1}
  • \dfrac {n-1}{3n-4}
If x=\sqrt 1+\dfrac{1}{{1}^{2}}+\dfrac{1}{{2}^{2}}+\sqrt 1+\dfrac{1}{{2}^{2}}+\dfrac{1}{{3}^{2}}+\sqrt 1+\dfrac{1}{{2019}^{2}}+\dfrac{1}{{2020}^{2}} then 4040(2020-x)
  • 1
  • 2
  • -1
  • -2
The sum of the series 1+\frac { 1.3 }{ 6 } +\frac { 1.3.5 }{ 6.8 } +...\infty is 
  • 4
  • 0
  • \infty
  • { (4) }^{ \frac { 2 }{ 3 } }
The sum of the infinite series \cot { ^{ 1 } } \left( \dfrac { 7 }{ 4 } \right) +\cot { ^{ -1 }\left( \dfrac { 19 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 39 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 67 }{ 4 } \right) } +........\infty is:
  • \\ \dfrac { \pi }{ 4 } -\cot { ^{ -1 } } (3)
  • \\ \dfrac { \pi }{ 4 } -\tan { ^{ -1 } } (3)
  • \\ \dfrac { \pi }{ 4 } +\cot { ^{ -1 } } (3)
  • \\ \dfrac { \pi }{ 4 } +\tan { ^{ -1 } } (3)
40280625, 732375, 16275, 465, 18.6, 1.24,?
  • 0.248
  • 0.336
  • 0.424
  • 0.512
  • 0.639
Sum of the series  S = 1^{2}-2^{2}+3^{2}-4^{2}+......-2002^{2}+2003^{2} is
  • 2007006
  • 1005004
  • 2000506
  • none of these
In the sum 3+33+333+3333+.........2015 terms the number formed by taking the last four digits in that order is 
  • 6365
  • 6255
  • 6465
  • 6565
Sum of the series \dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+.....(infinite terms) is?
  • e
  • 2e
  • e^2
  • \infty
If a_{k}=\dfrac{1}{K(K+1)} for K=1, 2, 3,.....n, then \left(\sum_{k=1}^{n}{a_{k}}\right)^{2}=
  • \dfrac{n}{n+1}
  • \dfrac{n^{2}}{(n+1)^{2}}
  • \dfrac{n^{4}}{(n+1)^{4}}
  • \dfrac{n^{6}}{(n+1)^{6}}
\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\frac { { C }_{ 3 } }{ { C }_{ 2 } } +......+\dfrac { n{ C }_{ n } }{ { C }_{ n-1 } } equals 
  • { n2 }^{ n-1 }
  • \dfrac { { n2 }^{ n-1 } }{ { 2 }^{ n }-1 }
  • \dfrac { n(n+1) }{ 2 }
  • \dfrac { (n+1)(n+2) }{ 2 }
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