12th Grade > Mathematics
BINOMIAL THEOREM MCQs
Binomial Theorem
Total Questions : 74
| Page 8 of 8 pages
Answer: Option C. -> 9
:
C
16=nC6(213)n−6(3−13)6nCn−6(213)6(3−13)n−6
6−1=613(n−12)
n−12=−3
n=9
:
C
16=nC6(213)n−6(3−13)6nCn−6(213)6(3−13)n−6
6−1=613(n−12)
n−12=−3
n=9
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 A. -> 5
:
A
15C2r+2 =15Cr−2
But15C2r+2 =15C15−(2r+2) =15C13−2r
⇒15C13−2r =15Cr−2 ⇒ r = 5.
:
A
15C2r+2 =15Cr−2
But15C2r+2 =15C15−(2r+2) =15C13−2r
⇒15C13−2r =15Cr−2 ⇒ r = 5.
Answer: Option C. -> 1024
:
C
The coefficients of the last six terms of (1+x)11 are
11C0,11C1,11C2....11C5
Sum=11C0+11C1+11C2+11C3+11C4+11C5=1+11+55+165+330+462=1024
:
C
The coefficients of the last six terms of (1+x)11 are
11C0,11C1,11C2....11C5
Sum=11C0+11C1+11C2+11C3+11C4+11C5=1+11+55+165+330+462=1024
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email