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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{\left( {1 + x} \right)^n} are in A.P then n =

Explanation

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\Rightarrow 2b = a + c
Secondly, the coefficient of rth term in the expansion (1+x)n{\left( {1 + x} \right)^n} is equal to the nCr1^n{C_{r - 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{\left( {1 + x} \right)^n} are in A.P.
The coefficient of rth term in the expansion (1+x)n{\left( {1 + x} \right)^n} will be equal to nCr1^n{C_{r - 1}}.
So, we can calculate the coefficient of the 9th, 10th and 11th term by putting the value of r in nCr1^n{C_{r - 1}}.
The coefficient of 9th term =nC8{ = ^n}{C_8},
The coefficient of 10th term =nC9{ = ^n}{C_9} and
The coefficient of 11th term =nC10{ = ^n}{C_{10}}.
Here, we will use the formula of combination i.e.
nCr=n!nr!.r!{ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{n - r!.r!}}
Therefore,
nC8=n!n8!.8!{ \Rightarrow ^n}{C_8} = \dfrac{{n!}}{{n - 8!.8!}} …………….(1)
nC9=n!n9!.9!{ \Rightarrow ^n}{C_9} = \dfrac{{n!}}{{n - 9!.9!}} ……………(2)
nC10=n!n10!.10!{ \Rightarrow ^n}{C_{10}} = \dfrac{{n!}}{{n - 10!.10!}} …………..(3)

As 9th, 10th and 11th term coefficient in the expansion (1+x)n{\left( {1 + x} \right)^n} are in A.P i.e. nC9^n{C_9}, nC9^n{C_9} andnC10^n{C_{10}}. Therefore,
2nC9=nC8+nC10\Rightarrow {2^n}{C_9}{ = ^n}{C_8}{ + ^n}{C_{10}} …………..(4)
From (1), (2), (3) and (4)
2(n!n9!.9!)=n!n8!.8!+n!n10!.10!\Rightarrow 2(\dfrac{{n!}}{{n - 9!.9!}}) = \dfrac{{n!}}{{n - 8!.8!}} + \dfrac{{n!}}{{n - 10!.10!}}
By taking n!n! common from each side and by cancelling them we get,
2(1n9!.9!)=1n8!.8!+1n10!.10!\Rightarrow 2(\dfrac{1}{{n - 9!.9!}}) = \dfrac{1}{{n - 8!.8!}} + \dfrac{1}{{n - 10!.10!}}
By opening of the required factorial, we get,
2(1(n9).n10!.9.8!)=1(n8)(n9).n10!.8!+1n10!.10.9.8!\Rightarrow 2(\dfrac{1}{{(n - 9).n - 10!.9.8!}}) = \dfrac{1}{{(n - 8)(n - 9).n - 10!.8!}} + \dfrac{1}{{n - 10!.10.9.8!}}
By taking common factorial from each term we get,
2×1n10!.8!×1(n9).9=1n10!.8!(1(n8)(n9)+110.9)\Rightarrow 2 \times \dfrac{1}{{n - 10!.8!}} \times \dfrac{1}{{(n - 9).9}} = \dfrac{1}{{n - 10!.8!}}(\dfrac{1}{{(n - 8)(n - 9)}} + \dfrac{1}{{10.9}})
By cancelling common factorial from each side, we get,
2×1(n9).9=1(n8)(n9)+110.9\Rightarrow 2 \times \dfrac{1}{{(n - 9).9}} = \dfrac{1}{{(n - 8)(n - 9)}} + \dfrac{1}{{10.9}}

2×1(n9).9=1(n8)(n9)+190 2×1(n9).9=90+(n8)(n9)90(n8)(n9)  \Rightarrow 2 \times \dfrac{1}{{(n - 9).9}} = \dfrac{1}{{(n - 8)(n - 9)}} + \dfrac{1}{{90}} \\\ \Rightarrow 2 \times \dfrac{1}{{(n - 9).9}} = \dfrac{{90 + (n - 8)(n - 9)}}{{90(n - 8)(n - 9)}} \\\

By cancelling (n9)(n - 9) and 9 with 90 from both sides we get,
2=90+(n8)(n9)10(n8)\Rightarrow 2 = \dfrac{{90 + (n - 8)(n - 9)}}{{10(n - 8)}}
By taking denominator of each side to other side to eliminate it, we get,
2×10×(n8)=90+(n8)(n9)\Rightarrow 2 \times 10 \times (n - 8) = 90 + (n - 8)(n - 9)
By opening the bracket, we get,

20n160=90+n28n9n+72 20n160=n217n+162  \Rightarrow 20n - 160 = 90 + {n^2} - 8n - 9n + 72 \\\ \Rightarrow 20n - 160 = {n^2} - 17n + 162 \\\

Taking all terms on one side and 0 on other we get,
n217n20n+162+160=0\Rightarrow {n^2} - 17n - 20n + 162 + 160 = 0
n237n+322=0\Rightarrow {n^2} - 37n + 322 = 0
We can find the value of n by factorization method.
n223n14n+322=0\Rightarrow {n^2} - 23n - 14n + 322 = 0
(n23)(n14)=0\Rightarrow (n - 23)(n - 14) = 0
n=14,23\Rightarrow 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^n{C_r} instead of nCr1^n{C_{r - 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.