24. Find n and x in the expansion of (1+x)", if the fifth term is four times the fourth term and the fourth term is 6 times the third term
Answers
Step-by-step explanation:
(1+x)
n
3rd term=
n
C
2
x
n−3
5th term=
n
C
4
x
n−5
n
C
2
x
n−3
=4
n
C
4
x
n−5
⇒
(n−2)!2!
n!
x
2
=
(x−4)!4!
4n!
⇒
(n−2)(n−3)(n−4)!2!
x
2
=
(x−4)!4×3×2!
4
⇒3x
2
=(n−2)(n−3)
⇒3x
2
=n
2
−5n+6---------------(1)
6thterm
4thterm
=
3
40
⇒
n
C
5
x
n−6
n
C
3
x
n−4
=
3
40
⇒
(n−3)(n−4)(n−5)!3!
5×4×3!×(n−5)!x
2
=
3
40
⇒3x
2
=2(n−3)(n−4)
⇒3x
2
=2(n−3)(n−4)
⇒3x
2
=2n
2
−14n+24-------------(2)
From Equation 1& 2,
n
2
−5n+6=2n
2
−14n+24
⇒n
2
−9n+18=0
⇒n
2
−6n−3n+18=0
⇒(n−6)(n−3)=0
⇒n=3 or n=6.
n
=3, then x=0
From, Equation 1
3x
2
=n
2
−5n+6
⇒3x
2
=6
2
−6⋅5+6o,
⇒3x
2
=36−30+6
⇒3x
2
=12
⇒x
2
=4
⇒x=±2
So, n=3,x=0
and n=6,x=±2
Step-by-step explanation:
Step-by-step explanation:
(1+x)
n
3rd term=
n
C
2
x
n−3
5th term=
n
C
4
x
n−5
n
C
2
x
n−3
=4
n
C
4
x
n−5
⇒
(n−2)!2!
n!
x
2
=
(x−4)!4!
4n!
⇒
(n−2)(n−3)(n−4)!2!
x
2
=
(x−4)!4×3×2!
4
⇒3x
2
=(n−2)(n−3)
⇒3x
2
=n
2
−5n+6---------------(1)
6thterm
4thterm
=
3
40
⇒
n
C
5
x
n−6
n
C
3
x
n−4
=
3
40
⇒
(n−3)(n−4)(n−5)!3!
5×4×3!×(n−5)!x
2
=
3
40
⇒3x
2
=2(n−3)(n−4)
⇒3x
2
=2(n−3)(n−4)
⇒3x
2
=2n
2
−14n+24-------------(2)
From Equation 1& 2,
n
2
−5n+6=2n
2
−14n+24
⇒n
2
−9n+18=0
⇒n
2
−6n−3n+18=0
⇒(n−6)(n−3)=0
⇒n=3 or n=6.
n
=3, then x=0
From, Equation 1
3x
2
=n
2
−5n+6
⇒3x
2
=6
2
−6⋅5+6o,
⇒3x
2
=36−30+6
⇒3x
2
=12
⇒x
2
=4
⇒x=±2
So, n=3,x=0
and n=6,x=±2