Question
Question: If the second, third and fourth terms in the expansion of \({\left( {x + a} \right)^n}\) are 240, 72...
If the second, third and fourth terms in the expansion of (x+a)n are 240, 720, 1080 respectively, then the value of n is
A) 15
B) 20
C) 10
D) 5
Solution
Hint- Here we will proceed by using the concept of binomial expansion that is (x+y)n=k=0∑n(kn)xn−kyk=k=0∑n(kn)xkyn−k.
Complete step-by-step answer:
We know that general term of (a+b)n is
Tr+1=nCr(a)n+r(b)r….. (1)
Given that second term of (x+b)n is 240
That is, T2=240 T1+1=240
Putting r=1, a=x and b=a
Tr+1=nCr(x)n−1(a)1
⇒T2=nC1xn−1a
⇒240=nC1xn−1a ….. (2)
Third term of (x+a)n is 720
That is T3=720
⇒T2+1=720
Putting r=2, a=x and b=a in (1)
T2+1=nC2(x)n−2.(b)2
⇒720=nC2xn−2a2 ….. (3)
Fourth term of (x+a)n is 1080
That is T4=1080
T3+1=1080
Putting r=3, a=x and b=a in (1)
T3+1=nCrxn−3a3
⇒1080=nC3xn−3a3 …. (4)
Also,
Dividing (3)by (2)
240720=nC1xn−1anC2xn−2a2
⇒3=nC1nC2×xn−1xn−2×aa2 ⇒3=1!(n−1)!n!2!(n−2)!n!×xn−2−(n−1)×a
⇒3=2!(n−2)!n!×n!1!(n−1)!×xn−2−n+1×a
⇒3=2(n−2)!n!×n!1(n−1)(n−2)!×x−1×a
⇒3=2(n−1)×xa
By cross multiplication, we will get
⇒3×2=(n−1)×xa
⇒6=(n−1)×xa
⇒n−16=xa ….. (A)
Now dividing (4) and (3)
⇒7201080=nC2xn−2a2nC3xn−3a3
⇒23=nC2nC3×xn−2xn−3×a2a3
⇒23=2(n−2)!n!3!(n−3)!n!×xn−3−n−2×a3−2 ⇒23=3!(n−3)!n!×n!2!(n−2)!×x−1×a1
⇒23=3(2)!(n−3)!n!×n!2!(n−2)(n−3)!×xa ⇒23=3(n−2)×xa
⇒2(n−2)9=xa …. (B)
Now our equations are
n−16=xa …. (A)
2(n−2)9=xa ….. (B)
Equating (A) and (B)
⇒2(n−2)9=n−16
By cross multiply, we will get
⇒9×(n−1)=6×2(n−2)
⇒9n−9=12n−24
⇒15=3n ⇒315=n ⇒5=n ⇒n=5
Hence, option D is the correct answer.
Note- In this question it should be noted that we use the basics of binomial expansion, elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. Also this theorem specifies the expansion of any power (a+b)m of a binomial (a+b) as a certain sum of products aibi, such as (a+b)2=a2+b2+2ab. By this method we can easily solve this question.