Processing math: 64%

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 4 - MCQExams.com

The sum of the series 1222+3242+..........10002+10012 is
  • 501×1001
  • 501×1001
  • 499×1001
  • 499×1001
1.22+2.32+3.42+....+n(n+1)212.2+22.3+32.4++n2(n+1)=
  • 3n+13n+5
  • 3n+53n+1
  • (3n+1)(3n+5)
  • 3n+53n+7
 12+1+22+2+32+3+..+n2+n=
  • (n+1)(n+2)3
  • n(n+1)(n+2)3
  • n(n+2)(n+3)6
  • (n+1)(n+2)(n+3)4
The sum of 7 terms of the series 1222+3242+5262+... is 
  • 21
  • 15
  • 28
  • 35
13+12+1+23+22+2+33+32+3+...3n terms = 
  • n(n+1)(n2+12n+5)12
  • n(n+1)(3n2+7n+8)12
  • n(n+1)(n+2)(n2+5n+6)12
  • (n+1)(n+2)(n+3)4
The sum V1+V2++Vn is 

  • 112n(n+1)(3n2n+1)
  • 112n(n+1)(3n2+n+2)
  • 12n(2n2n+1)
  • 13(2n32n+3)
The sum of the infinite series, 11.3+23.7+37.13+413.21+.......... is

  • 12
  • 2524
  • 2554
  • 125252
Find n=11(2n1)!
  • 12(e+e1)
  • 12(ee1)
  • e+e1
  • ee1
121+12+221+2+12+22+321+2+3++n terms =
  • n(n+3)4
  • n(n+3)5
  • n(n+2)3
  • n(n+5)6
Evaluate 1.6+2.9+3.12+...+n(3n+3)=
  • n(n+1)(n+2)
  • (n+1)(n+2)(n+3)
  • (n+2)(n+3)(n+4)
  • (n1)n(n+1)
2.4+4.7+6.10+...(n1) terms
  • 2n32n2
  • n3+3n2+16
  • 2n3+2n
  • 2n3n2
The sum of the first n terms of the series 12+2.22+32+2.42+52+2.62+..... is n(n+1)22, when n is even,when n is odd, the sum is
  • n2(n+1)2
  • [n(n+1)2]2
  • 3n(n+1)2
  • n(n+1)24
x=(1)a1+(1)a2+.....(1)a1006
y=(1)a1007+(1)a1008+...+(1)a2013
Then which of the following is true?
  • (1)x=1;(1)y=1
  • (1)x=1;(1)y=1
  • (1)x=1;(1)y=1
  • (1)x=1;(1)y=1
What number corresponds to the fifth missing planet ?
  • 28
  • 78
  • 32
  • 36
State true or false:
Given sequence 15,30,60,120,240 is in G.P.
  • True
  • False
Let a sequence {an} be defined by an=1n+1+1n+2+1n+3+.....+13n, then
  • a2=712
  • a2=1920
  • an+1an=9n+5(3n+1)(3n+2)(3n+3)
  • an+1an=23(n+1)
In the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, .......... starting with the third terms, each term is the sum of the two preceding terms : 2=1+1,3=1+2,5=2+3, and so on.
Choose the correct options -
  • Eighth term in Fibonacci sequence is 21
  • Tenth term in Fibonacci sequence is 55
  • Twelfth term in Fibonacci sequence is 144
  • Twelfth term in Fibonacci sequence is 112
State true or false:
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Parts of red pigment
1
4
7
12
20
Parts of base
8
...
...
...
...

Answer: 
Parts of red pigment
1
4
7
12
20
Parts of base
8
325696150
  • True
  • False
The value of α for which n=1An=83 is-
114347_20b59845404741bc87d02969783237ad.png
  • 1/3, 2/3
  • 1/4, 3/4
  • 1/5, 4/5
  • 1/2
Out of nine cells of a square, one cell is left blank and in the rest of the cells, numbers are written follow some rule. Get the rule and find out the proper option for the blank cell 
2
72
56
?
0
42
12
20
30
  • 4
  • 6
  • 8
  • 10
Which of the following is not a G.P.?
  • 2,4,6,8....
  • 5,25,125,625....
  • 1.5,3.0,6.0,12.0....
  • 8,16,24,32,....
The value of 10n=1n0xdx is
  • an even integer
  • an odd integer
  • a rational number
  • an irrational number
If n is an odd integer greater than or equal to 1 then the value of n3(n1)3+(n2)3...+(1)n1.13 is
  • (n+1)2.(2n1)4
  • (n1)2.(2n1)4
  • (n+1)2.(2n+1)4
  • none of these
Find the sum of first n terms of the series.
13+3×22+33+3×42+53+3×62+.... If n is even.
  • n8(n3+4n2+10n+6)
  • n8(n2+4n2+10n+6)
  • n8(n3+4n2+10n+8)
  • n8(n2+4n2+10n+8)
Let Sn denote the sum of cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then nr=1SrSr is equal to 
  • n(n+1)(n+2)6
  • n(n+1)2
  • n2+3n+26
  • none of these
Let a sequence be defined by a1=0 and an+1=an+4n+3 for all n1(nϵN)

The value of ak in terms of k is (kN)

  • (k1)(2k+3)
  • (k1)(3k+1)
  • (k1)(k+3)
  • (k1)(2k+5)
Let Vr denote the sum of the first r terms of an arithmetic progression (AP) whose first term is r and the common difference is (2r1). Let  Tr=Vr+1Vr2 and  Qr=Tr+1Tr for r=1,2,...
The sum  V1+V2+...+Vn is
  • 112n(n+1)(3n2n+1)
  • 112n(n+1)(3n2+n+2)
  • 12n(2n2n+1)
  • 13(2n32n+3)
Find the sum of the products of every pair of the first n natural numbers.
  • S=(n1)2(3n+2)24
  • S=n(n+1)(n1)(3n2)24
  • S=n(n1)2(3n+2)24
  • S=n(n+1)(n1)(3n+2)24
The value of nr=1{(2r1)a+1br} is equal to 
  • an2+bn11bn1(b1)
  • an2+bn1bn(b1)
  • an3+bn11bn(b1)
  • none of these
Sum of the series 11+23+45+87... up to n terms
  • 10(2n1)+n2
  • 10(2n+1)+n2
  • 20(2nn)+n2
  • None of these
12+(12+22)+(12+22+32)+... to n terms.
  • 112n(n+1)(2n3+3n2+n)
  • 112n(n+1)(3n3+2n2+n)
  • 16n(n+1)(2n+1)
  • 14n2(n+1)2
1323+3343+...+93 equals
  • 425
  • 400
  • 405
  • 395
The sum of all possible product of 1st n natural numbers taken two at a time is
  • n(n+1)(n+2)(2n+3)24
  • n(n21)(3n+2)24
  • n(n2+1)(3n+2)6
  • n(n21)(3n+2)6
Sum of the series 2.3.1+3.4.4+4.5.7+... up to n terms is
  • n12[9n3+46n2+51n34]
  • n6[9n3+46n2+51n34]
  • (n+1)6[9n3+46n234]
  • \displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}-34 \right ]
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.
  • \displaystyle n\geq 1.
  • \displaystyle n\ne 1.
  • \displaystyle n= 1.
  • None of the above
  • Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
  • Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
  • Assertion is correct but Reason is incorrect
  • Both Assertion and Reason are incorrect
\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
  • \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)
The sum of n terms of 1^{2}\, +\, (1^{2}\, +\, 2^{2})\, +\, (1^{2}\, +\, 2^{2}\, +\, 3^{2})\, +\, .....
  • \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}
1 + 3 + 6 + 10 + .... upto n terms is equal to
  • \displaystyle \frac{1}{3} n(n + 1)(n + 2)
  • \displaystyle \frac{1}{6} n(n + 1)(n + 2)
  • \displaystyle \frac{1}{12} n(n + 2)(n + 3)
  • \displaystyle \frac{1}{12} n(n + 1)(n + 2)
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 ).
  • \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 ).
The sum of n terms of the series
1 + (1 + a) + (1 + a + a^{2}) + (1 + a + a^{2} + a^{3}) + ....., is
  • \displaystyle \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}
  • \displaystyle \frac{n}{1 - a} + \frac{a(1 - a^{n})}{(1 - a)^{2}}
  • \displaystyle \frac{n}{1 - a} - \frac{a(1 + a^{n})}{(1 - a)^{2}}
  • \displaystyle - \frac {n}{1 - a} + \frac{a(1 - a^{n})}{(1 - a)^{2}}
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}
  • 999
  • Greatest  3 digit number
  • Either A or B
  • Smallest  3 digit number
Sum to 20 terms of the series 1.{ 3 }^{ 2 }+2.{ 5 }^{ 2 }+3.{ 7 }^{ 2 }+... is
  • 178090
  • 168090
  • 188090
  • None of these
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 )
  • \displaystyle \frac{5}{12}
  • \displaystyle \frac{1}{2}
  • \displaystyle \frac{11}{20}
  • \displaystyle \frac{7}{10}
If \displaystyle \sum_{s=1}^{n}\, \left \{ \displaystyle \sum_{r=1}^{s}r \right \}\, =\, an^3\, +\, bn^2\, +\, cn, then find the value of a + b + c.
  • 1
  • 0
  • 2
  • 3
The sum of the series 4 + 8 + 16 + 32 + ........ till 10 terms is
  • 6\, \times\, 2^{10}\, -\, 1
  • 4\, \times\, 2^{10}\, +\, 1
  • 4(2^{10}\, -\, 1)
  • 4(2^{10}\, +\, 1)
\displaystyle \sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi  }{ 11 }  } -i\cos { \frac { 2q\pi  }{ 11 }  }  \right)  }  \right]  }^{ p } } =
  • 8\left( 1-i \right)
  • 16\left( 1-i \right)
  • 48\left( 1-i \right)
  • None of these
Find the sum the infinite G.P.:
1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}
  • \displaystyle \frac{3}{5}
  • \displaystyle \frac{3}{2}
  • \displaystyle \frac{49}{27}
  • \displaystyle \frac{8}{5}
The sum of series \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
  • \displaystyle\frac {1}{3}(2n+1)
  • \displaystyle\frac {1}{3}n^2
  • \displaystyle\frac {1}{3}(n+2)
  • \displaystyle\frac {1}{3}n(n+2)
\dfrac {1^3+2^3+3^3+4^3+........12^3}{1^2+2^2+3^2+4^2+......12^2}=
  • \dfrac {234}{25}
  • \dfrac {243}{35}
  • \dfrac {263}{27}
  • None of these
0:0:2


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 11 Engineering Maths Quiz Questions and Answers