12th Grade > Mathematics
BINOMIAL THEOREM MCQs
Binomial Theorem
Total Questions : 74
| Page 6 of 8 pages
Answer: Option C. -> n(n+1)2
:
C
C1C0+2⋅C2C1+3⋅C3C2+........+n⋅CnCn−1
=n1+2n(n−1)1.2n+3n(n−1)(n−2)3.2.1n(n−1)1.2+......+n.1n
=n+(n−1)+(n−2)........+1=∑n=n(n+1)2
:
C
C1C0+2⋅C2C1+3⋅C3C2+........+n⋅CnCn−1
=n1+2n(n−1)1.2n+3n(n−1)(n−2)3.2.1n(n−1)1.2+......+n.1n
=n+(n−1)+(n−2)........+1=∑n=n(n+1)2
Answer: Option B. -> (r+1)(r+2)(r+3)6xr
:
B
(1−x)−4 = [1.2.36x0+2.3.46x+3.4.56x2+4.5.66x3+.......+(r+1)(r+2)(r+3)6xr+.........]
Therefore Tr+1 = (r+1)(r+2)(r+3)6xr.
:
B
(1−x)−4 = [1.2.36x0+2.3.46x+3.4.56x2+4.5.66x3+.......+(r+1)(r+2)(r+3)6xr+.........]
Therefore Tr+1 = (r+1)(r+2)(r+3)6xr.
Answer: Option B. -> n2−n(4p+1)+4p2−2 = 0
:
B
2ncp=ncp−1+pcp+1⇒2(n−p)!P!=1(n−p+1)!(p−1)!+1(n−p−1)!(p+1)!⇒2(n−p)p=1(n−p+1)(n−p)+1(p+1)ponsimplifingwegetn2−n(4p+1)+4p2−2=0
:
B
2ncp=ncp−1+pcp+1⇒2(n−p)!P!=1(n−p+1)!(p−1)!+1(n−p−1)!(p+1)!⇒2(n−p)p=1(n−p+1)(n−p)+1(p+1)ponsimplifingwegetn2−n(4p+1)+4p2−2=0
Answer: Option C. -> 12a11b
:
C
Tr+1=12cr(ax)12−r(bx)r=12cr.a12−r.br.x2r−12∴2x−12=−10⇒r=1∴coefficient=12.a11.b
:
C
Tr+1=12cr(ax)12−r(bx)r=12cr.a12−r.br.x2r−12∴2x−12=−10⇒r=1∴coefficient=12.a11.b
Answer: Option D. -> 1.3.5.......(2n−1)n!2nxn
:
D
Middle term of (1+2x)2n is Tn+1 =2nCnxn
= (2n)!n!n!xn = 1.3.5.......(2n−1)n! 2nxn.
:
D
Middle term of (1+2x)2n is Tn+1 =2nCnxn
= (2n)!n!n!xn = 1.3.5.......(2n−1)n! 2nxn.
Answer: Option C. -> p = q + r
:
C
Since (n+2)th term is the middle term in the
expansion of (1+x)2n+2, therefore p =2n+2Cn+1.
Since (n+1)th and (n+2)th terms are middle terms in the expansion of (1+x)2n+1,
thereforeq =2n+1Cn and r =2n+1C2n+1.
But ,2n+1Cn +2n+1Cn+1 =2n+2Cn+1
∴ q + r = p
:
C
Since (n+2)th term is the middle term in the
expansion of (1+x)2n+2, therefore p =2n+2Cn+1.
Since (n+1)th and (n+2)th terms are middle terms in the expansion of (1+x)2n+1,
thereforeq =2n+1Cn and r =2n+1C2n+1.
But ,2n+1Cn +2n+1Cn+1 =2n+2Cn+1
∴ q + r = p
Answer: Option C. -> 3
:
C
Tr+1 =5Cr(x2)5−r(kx)r
For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3
Hence, T3+1 =5C3(x2)5−3(kx)3
According to question, 10k3 = 270 ⇒ k = 3.
:
C
Tr+1 =5Cr(x2)5−r(kx)r
For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3
Hence, T3+1 =5C3(x2)5−3(kx)3
According to question, 10k3 = 270 ⇒ k = 3.
Answer: Option C. -> 12a11b
:
C
Tr+1=12cr(ax)12−r(bx)r=12cr.a12−r.br.x2r−12∴2x−12=−10⇒r=1∴coefficient=12.a11.b
:
C
Tr+1=12cr(ax)12−r(bx)r=12cr.a12−r.br.x2r−12∴2x−12=−10⇒r=1∴coefficient=12.a11.b
Answer: Option A. -> 1
:
A
In the expansion of (ax2+1bx)11, the general
term is
Tr+1 =11Cr(ax2)11−r(1bx)r =11Cr(a)11−r(1br)x22−3r
For x7, we must have 22 - 3r = 7 ⇒ r = 5, and
the coefficient of x7 =11C5.a11−515) =11C5 a6b5
Similarly, in the expansion of (ax−1bx2)11, the
general term is Tr+1 =11Cr(−1)r a11−rbr.x11−3r
For x−7 we must have, 11 - 3r = -7 ⇒ r = 6, and
the coefficient of x−7 is11C6 a5b6 =11C5 a5b6.
As given,11C5 a6b5 =11C5 a5b6⇒ ab = 1.
:
A
In the expansion of (ax2+1bx)11, the general
term is
Tr+1 =11Cr(ax2)11−r(1bx)r =11Cr(a)11−r(1br)x22−3r
For x7, we must have 22 - 3r = 7 ⇒ r = 5, and
the coefficient of x7 =11C5.a11−515) =11C5 a6b5
Similarly, in the expansion of (ax−1bx2)11, the
general term is Tr+1 =11Cr(−1)r a11−rbr.x11−3r
For x−7 we must have, 11 - 3r = -7 ⇒ r = 6, and
the coefficient of x−7 is11C6 a5b6 =11C5 a5b6.
As given,11C5 a6b5 =11C5 a5b6⇒ ab = 1.
Answer: Option D. -> (16,2723)
:
D
(1+ax+bx2)(1–2x)18=1(1−2x)18+ax(1−2x)18+bx2(1−2x)18
Coefficient of x3:(−2)318C3+a(−2)218C2+b(−2)18C1=0
4×(17×16)(3×2)−2a×172+b=0----- (1)
Coefficient of x4:(−2)418C4+a(−2)318C3+b(−2)218C2=0
(4×20)−2a×163+b=0 ---- (2)
From equations (1) and (2), we get
4(17×83−20)+2a(163−172)=0
⇒ a = 16
⇒b=2×16×163−80=2723
:
D
(1+ax+bx2)(1–2x)18=1(1−2x)18+ax(1−2x)18+bx2(1−2x)18
Coefficient of x3:(−2)318C3+a(−2)218C2+b(−2)18C1=0
4×(17×16)(3×2)−2a×172+b=0----- (1)
Coefficient of x4:(−2)418C4+a(−2)318C3+b(−2)218C2=0
(4×20)−2a×163+b=0 ---- (2)
From equations (1) and (2), we get
4(17×83−20)+2a(163−172)=0
⇒ a = 16
⇒b=2×16×163−80=2723
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