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Question: The coefficients of three consecutive terms of \[{\left( {1 + x} \right)^{n + 5}}\] are in the ratio...

The coefficients of three consecutive terms of (1+x)n+5{\left( {1 + x} \right)^{n + 5}} are in the ratio 5:10:145:10:14, then n=n =

Explanation

Solution

In this question, we will proceed by letting the consecutive terms and finding their coefficients. Then use the given ratio two find two equations in terms of nn. Solve them to find the required answer.

Complete step by step answer:
Let the three consecutive terms be (r−1)th,rth{\left( {r - 1} \right)^{th}},{r^{th}} and (r+1)th{\left( {r + 1} \right)^{th}} terms i.e., Tr−1,Tr,Tr+1{T_{r - 1}},{T_r},{T_{r + 1}} of the expansion (1+x)n+5{\left( {1 + x} \right)^{n + 5}}.
We know that the general term of expansion (1+x)n{\left( {1 + x} \right)^n} is Tr+1=nCrxr{T_{r + 1}} = {}^n{C_r}{x^r}
So, for the expansion (1+x)n+5{\left( {1 + x} \right)^{n + 5}} we have

⇒Tr−1=n+5Cr−2xr−2 ⇒Tr=n+5Cr−1xr−1 ⇒Tr+1=n+5Crxr  \Rightarrow {T_{r - 1}} = {}^{n + 5}{C_{r - 2}}{x^{r - 2}} \\\ \Rightarrow {T_r} = {}^{n + 5}{C_{r - 1}}{x^{r - 1}} \\\ \Rightarrow {T_{r + 1}} = {}^{n + 5}{C_r}{x^r} \\\

Hence the coefficients of terms (r−1)th,rth{\left( {r - 1} \right)^{th}},{r^{th}} and (r+1)th{\left( {r + 1} \right)^{th}} are n+5Cr−2,n+5Cr−1{}^{n + 5}{C_{r - 2}},{}^{n + 5}{C_{r - 1}} and n+5Cr{}^{n + 5}{C_r} respectively.
Given that the coefficients of (r−1)th,rth{\left( {r - 1} \right)^{th}},{r^{th}} and (r+1)th{\left( {r + 1} \right)^{th}} terms are in a ration of 5:10:145:10:14.
So, we have

⇒coefficient of (r−1)thcoefficient of rth=510 ⇒n+5Cr−2n+5Cr−1=510  \Rightarrow \dfrac{{{\text{coefficient of }}{{\left( {r - 1} \right)}^{th}}}}{{{\text{coefficient of }}{r^{th}}}} = \dfrac{5}{{10}} \\\ \Rightarrow \dfrac{{{}^{n + 5}{C_{r - 2}}}}{{{}^{n + 5}{C_{r - 1}}}} = \dfrac{5}{{10}} \\\

Using the formula nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} we have

⇒(n+5)!(r−2)![n−(r−2)]!(n+5)!(r−1)![n−(r−1)]!=510 ⇒(r−1)![n−(r−1)]!(r−2)![n−(r−2)]!=510 ⇒(r−1)(r−2)![n−(r−1)]!(r−2)!(n−(r−2))[n−(r−1)]!=510 ⇒(r−1)n−(r−2)=510 ⇒10r−10=5n−5r+10 ⇒5n=15r−20....................................................(1)  \Rightarrow \dfrac{{\dfrac{{\left( {n + 5} \right)!}}{{\left( {r - 2} \right)!\left[ {n - \left( {r - 2} \right)} \right]!}}}}{{\dfrac{{\left( {n + 5} \right)!}}{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}}} = \dfrac{5}{{10}} \\\ \Rightarrow \dfrac{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}{{\left( {r - 2} \right)!\left[ {n - \left( {r - 2} \right)} \right]!}} = \dfrac{5}{{10}} \\\ \Rightarrow \dfrac{{\left( {r - 1} \right)\left( {r - 2} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}{{\left( {r - 2} \right)!\left( {n - \left( {r - 2} \right)} \right)\left[ {n - \left( {r - 1} \right)} \right]!}} = \dfrac{5}{{10}} \\\ \Rightarrow \dfrac{{\left( {r - 1} \right)}}{{n - \left( {r - 2} \right)}} = \dfrac{5}{{10}} \\\ \Rightarrow 10r - 10 = 5n - 5r + 10 \\\ \Rightarrow 5n = 15r - 20....................................................\left( 1 \right) \\\

Also, we have

⇒coefficient of rthcoefficient of (r+1)th=1014 ⇒n+5Cr−2n+5Cr−1=57  \Rightarrow \dfrac{{{\text{coefficient of }}{r^{th}}}}{{{\text{coefficient of }}{{\left( {r + 1} \right)}^{th}}}} = \dfrac{{10}}{{14}} \\\ \Rightarrow \dfrac{{{}^{n + 5}{C_{r - 2}}}}{{{}^{n + 5}{C_{r - 1}}}} = \dfrac{5}{7} \\\

Using the formula nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} we have

⇒(n+5)!(r−1)![n−(r−1)]!(n+5)!(r)![n−(r)]!=57 ⇒(r)![n−(r)]!(r−1)![n−(r−1)]!=57 ⇒(r)(r−1)![n−(r)]!(r−1)!(n−(r−1))[n−(r)]!=57 ⇒(r)n−(r−1)=57 ⇒7r=5n−5r+5 ⇒5n−12r+5=0 ⇒5n=12r−5....................................................(2)  \Rightarrow \dfrac{{\dfrac{{\left( {n + 5} \right)!}}{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}}}{{\dfrac{{\left( {n + 5} \right)!}}{{\left( r \right)!\left[ {n - \left( r \right)} \right]!}}}} = \dfrac{5}{7} \\\ \Rightarrow \dfrac{{\left( r \right)!\left[ {n - \left( r \right)} \right]!}}{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}} = \dfrac{5}{7} \\\ \Rightarrow \dfrac{{\left( r \right)\left( {r - 1} \right)!\left[ {n - \left( r \right)} \right]!}}{{\left( {r - 1} \right)!\left( {n - \left( {r - 1} \right)} \right)\left[ {n - \left( r \right)} \right]!}} = \dfrac{5}{7} \\\ \Rightarrow \dfrac{{\left( r \right)}}{{n - \left( {r - 1} \right)}} = \dfrac{5}{7} \\\ \Rightarrow 7r = 5n - 5r + 5 \\\ \Rightarrow 5n - 12r + 5 = 0 \\\ \Rightarrow 5n = 12r - 5....................................................\left( 2 \right) \\\

From equation (1) and (2), we have

⇒15r−20=12r−5 ⇒15r−12r=20−5 ⇒3r=15 ⇒r=153 ∴r=5  \Rightarrow 15r - 20 = 12r - 5 \\\ \Rightarrow 15r - 12r = 20 - 5 \\\ \Rightarrow 3r = 15 \\\ \Rightarrow r = \dfrac{{15}}{3} \\\ \therefore r = 5 \\\

Substituting r=5r = 5 in equation (1), we have

⇒5n=15(5)−20 ⇒5n=75−20 ⇒5n=55 ⇒n=555 ∴n=11  \Rightarrow 5n = 15\left( 5 \right) - 20 \\\ \Rightarrow 5n = 75 - 20 \\\ \Rightarrow 5n = 55 \\\ \Rightarrow n = \dfrac{{55}}{5} \\\ \therefore n = 11 \\\

Thus, the value of nn is 11.

Note:
The general term of expansion (1+x)n{\left( {1 + x} \right)^n} is Tr+1=nCrxr{T_{r + 1}} = {}^n{C_r}{x^r}. Don’t confuse when you are expanding the terms of combinations. We can also check our answer by substituting the values of r,nr,n in the considered terms by checking whether their coefficients are consecutive or not.