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CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 3 - MCQExams.com

Find the sum nr=1rnCrCr1
  • n(n+1)3!
  • n(n1)2
  • n(n+1)2
  • None of these
If the middle term in the expansion of (x2+1x)n is 924x6, then find the value of n
  • n=11
  • n=13
  • n=12
  • n=14
The middle term in the expansion of (xy+yx)8 is.
  • 8C5
  • 8C6
  • 8C4
  • 8C2
If the coefficient of (2r + 4)th term is equal to the coefficient of (r - 2)th term in the expansion of (1+x)18 then r=
  • 2
  • 4
  • 6
  • 8
If the coeffecient of the middle term in the expansion of (1+x)2n+2 is α and coeffecient of middle terms in the expansion of (1+x)2n+1 are β and γ, then relate α,β and γ
  • βγ=α
  • γβ=α
  • β+γ=α
  • none of these
If the coefficients of x7 and x8 in (2+x3)n are equal then n =
  • 45
  • 55
  • 35
  • 27
The middle term in the expansion of (1+x)2n is
  • 2nxn(135(2n1))n!
  • 2nxn(135(2n1))n!
  • 2nxn(135(2n1))n1
  • 2nxn(135(2n1))n1
In the second term in the expansion (13a+aa1)n is 14a52 , then the value of nC3nC2 is
  • 8
  • 12
  • 4
  • None of these
The coeffecients of the middle term in the binomial expansion in powers of x of (1+αx)4 and (1+αx)6 is the same if α equals
  • 53
  • 103
  • 310
  • 35
The middle term in the expansion of (x/2+2)8 is 1120, then xεR is equal to
  • 2
  • 3
  • 3
  • 2
If C0,C1,C2,...,Cn are the coefficients of the expansion of (1+x)n, then the value of n0Ckk+1 is
  • 0
  • 2n1n
  • 2n+11n+1
  • None of these
The middle term in the expansion of (11x)n(1x)n, is
  • 2nCn
  • 2nCn
  • 2nCn1
  • none of these
If A is the coefficient of the middle term in the expansion of (1+x)2n and B and C are the coefficients of two middle terms in the expansion of (1+x)2n1, then
  • A+B=C
  • B+C=A
  • C+A=B
  • A+B+C=0
If n is an integer between 0 and 21, then the  minimum value of n!(21n)! is attained for n=
  • 1
  • 10
  • 12
  • 20
The middle term in the expansion of(1+x)2nis
  • 135...(2n1)n!2nxn
  • 135...(2n1)n!2nxnxn
  • 2nCn
  • 2nCn1xn1
If the coefficient of x7 in [ax+(1bx)]11is 55a11, then a and b satisfy the relation   
  • a+b=1
  • ab=1
  • ab=1
  • ab=1
The middle term in the expansion of (x+1x)10,is
  • 10C11x
  • 10C5
  • 10C6
  • 10C7x
The value of nC12+nC34+nC56+... is
  • 2n1n
  • 2n+1n
  • 2n1n+1
  • 2n+1n+1
Find the sum 1 × 2 × C1 + 2 × 3 C2 + n (n+1)Cn where Cr = nCr
  • n(n+1)2n1
  • n(n+3)2n2
  • 2n(2nCn)
  • None of these
If(1+2x+x2)n=2nr=0arxr, then ar=
  • (nCr)2
  • nCrnCr+1
  • 2nCr
  • 2nCr+1
Find the ratio of the coefficient of x10 in (1x2)10 and the term independent of x in the expansion of (x2x)10
  • 1:8
  • 1:16
  • 1:24
  • 1:32
Find the coefficient of 1y2 in (y+cy2)10.
  • 210c4
  • 210c5
  • 120c3
  • 120c4
The Value of nC1+n+1C2+n+2C3+...+n+m1Cm is equal 
  • m+nCn1
  • m+nCn+1
  • mC1+m+1C2+m+2C3+...+n+m1Cn
  • m+nCm1
In the expansion of (713+1119)6561,
  • there are exactly 730 rational terms
  • there are exactly 5832 irrational terms
  • the term which involves greatest binomial coefficient is irrational
  • the term which involves greatest binomial coefficients is rational
If (1+x)n=C0+C1x+C2x2+...+Cnxn, then 2C0+22.C12+23.C23+...+2n+1.Cnn+1=
  • 3n+11n+1
  • 3n1n
  • 3n+21n+2
  • None of these
Determine the value of x in the expression of (2+x)5, if the second term in the expansion is 240
  • (24)14
  • 6
  • 3
  • None of the above
Find the (n+1)th term from the end in the expansion of (x1x)2n
  • (1)n.2nCn
  • (1)n+1.2nCn1
  • (1)n.2nCn1
  • None of these
Find the middle term in the expansion of (2x3+32x)10.
  • 210
  • 630
  • 252
  • 756
The fourth term in the expansion of (px+1x)n is 52. Then,
  • n=6
  • n=7
  • p=12
  • p=14
If (1+2x+x2)n=2nr=0arxr, then ar=
  • (nCr)2
  • nCr.nCr+1
  • 2nCr
  • 2nCr+1
The middle term in the expansion of (1+x)2n is , n being a positive integer is
  • {1.3.5....(2n)}2nn!xn
  • {1.3.5....(2n)}2nn!n!xn
  • {1.3.5....(2n1)}2nn!n!xn
  • {1.3.5....(2n1)}2nn!xn
If n>2, then find the value of C1(a1)2C2(a2)2+C3(a3)2.....+(1)n1Cn(an)2 where Cr stands for nCr
  • a3
  • a
  • a2
  • a2
find the 7th term in the expansion of (4x12x)13
  • 439296x7
  • 439296x4
  • 439396x7
  • 43396x4
The 4th term from the end in the expansion of (x322x2)7 is

  • 35x
  • 70x2
  • 35x2
  • 70x
The middle term in the expansion of (ax+bx)12 is
  • 924a6b6
  • 924a6b5
  • 924a5b5
  • 924a5b6
The 8th term of (3x+23x2)12, when expanded ina scending power of x, is
  • 228096x3
  • 228096x9
  • 328179x3
  • None of these
Find the middle term in the expansion of (3xx36)9.

  • 9C6(3x)5
  • 9C5(3x)4
  • Both A & B
  • none of the above
If (8+37)n=α+β  where n and α are positive integers and β is a positive proper fraction,then
  • (1β)(α+β)=1
  • (1+β)(α+β)=1
  • (1β)(αβ)=1
  • (1+β)(αβ)=1
If (1+x)n=C0+C1x+C2x2+...+Cnxn, then 0ijn(Ci+Cj)2=
  • (n1).2nCn+22n
  • n.2nCn+22n
  • (n+1).2nCn+22n
  • None of these
Find the middle term in the expansion of (xaax)21
  • 20C10xa,21C10ax
  • 20C9xa,21C10ax
  • 21C10xa,21C10ax
  • 21C9xa,21C10ax
If the second term in the expansion [a113+aa1]n is 14 a5/2, then the value of nC3nC2 is
  • 4
  • 3
  • 12
  • 6
If the number of terms in (x+1+1x)n(nI+) is 401, then n is greater than
  • 201
  • 200
  • 199
  • none of these
If an=nr=01nCrthen nr=0rnCr equals
  • (n1)an
  • nan
  • 12nan
  • (n1)2an
The total number of terms in the expansion of (x+a)100+(xa)100  after simplification is

  • 202
  • 51
  • 50
  • 49
Let n and k be  positive integers such that nk(k+1)2. The number of solution (x1,x2,..,xk)1;x22,...,xkk all integers satisfying x1+x2+x3+...+xk=n is
  • mCk1
  • mCk3
  • mCk+1
  • None of these
The number of irrational terms in the expansion of (215+3110)55 is

  • 47
  • 56
  • 50
  • 48
If nCr+4nCr+1+6nCr+2+4nCr+3+nCr+4nCr+3nCr+1+3nCr+2+nCr+3=n+kr+k. Find the value of k
  • 2
  • 4
  • 6
  • 8
In the expansion of (x+x21)6+ (xx21)6,the number of terms is
  • 7
  • 14
  • 6
  • 4
The number of real negative terms in the binomial expansion of (1+ix)4n2, nϵN, x>0, is
  • n
  • n+1
  • n1
  • 2n
Find the sum of the series nr=0(1)nnCr[12r+3r22r+7r23r+15r24r+...uptomterms]
  • (2mn+1)(2n+1)(2mn)
  • (2mn1)(2n1)(2mn)
  • (2mn+1)(2n1)(2mn)
  • (2mn1)(2n+1)(2mn)
0:0:3


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