Question
Question: The coefficient of 9th, 10th and 11th term in the expansion of \({\left( {1 + x} \right)^n}\) are in...
The coefficient of 9th, 10th and 11th term in the expansion of (1+x)n are in A.P then n =
Solution
The numbers of a series are in A.P if the difference between two consecutive numbers are the same for all throughout the series. If a, b and c are in A.P then
⇒2b=a+c
Secondly, the coefficient of rth term in the expansion (1+x)n is equal to the nCr−1.
By relating these two concepts we can calculate the value of n.
Complete step-by-step answer:
In this question we have given that 9th, 10th and 11th term coefficient in the expansion (1+x)n are in A.P.
The coefficient of rth term in the expansion (1+x)n will be equal to nCr−1.
So, we can calculate the coefficient of the 9th, 10th and 11th term by putting the value of r in nCr−1.
The coefficient of 9th term =nC8,
The coefficient of 10th term =nC9 and
The coefficient of 11th term =nC10.
Here, we will use the formula of combination i.e.
⇒nCr=n−r!.r!n!
Therefore,
⇒nC8=n−8!.8!n! …………….(1)
⇒nC9=n−9!.9!n! ……………(2)
⇒nC10=n−10!.10!n! …………..(3)
As 9th, 10th and 11th term coefficient in the expansion (1+x)n are in A.P i.e. nC9, nC9 andnC10. Therefore,
⇒2nC9=nC8+nC10 …………..(4)
From (1), (2), (3) and (4)
⇒2(n−9!.9!n!)=n−8!.8!n!+n−10!.10!n!
By taking n! common from each side and by cancelling them we get,
⇒2(n−9!.9!1)=n−8!.8!1+n−10!.10!1
By opening of the required factorial, we get,
⇒2((n−9).n−10!.9.8!1)=(n−8)(n−9).n−10!.8!1+n−10!.10.9.8!1
By taking common factorial from each term we get,
⇒2×n−10!.8!1×(n−9).91=n−10!.8!1((n−8)(n−9)1+10.91)
By cancelling common factorial from each side, we get,
⇒2×(n−9).91=(n−8)(n−9)1+10.91
By cancelling (n−9) and 9 with 90 from both sides we get,
⇒2=10(n−8)90+(n−8)(n−9)
By taking denominator of each side to other side to eliminate it, we get,
⇒2×10×(n−8)=90+(n−8)(n−9)
By opening the bracket, we get,
Taking all terms on one side and 0 on other we get,
⇒n2−17n−20n+162+160=0
⇒n2−37n+322=0
We can find the value of n by factorization method.
⇒n2−23n−14n+322=0
⇒(n−23)(n−14)=0
⇒n=14,23
Therefore, the value of n can be 14 or 23.
Note: To calculate the coefficient of the given expansion some make mistakes i.e. they take nCr instead of nCr−1. If you take that then your answer will get wrong. Also, while seeing expansion what will some students do? They just consider n as a term and write their expansion. In this way they got struck.
n and r are different things. Here n means power of expansion and r means the term. Take care of this thing. They have said their coefficients are in A.P not the complete term.