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Question: If the coefficients of \( {\text{}}{{\text{r}}^{{\text{th}}}} \) , \( {{\text{(r + 1)}}^{{\text{th}}...

If the coefficients of rth{\text{}}{{\text{r}}^{{\text{th}}}} , (r + 1)th{{\text{(r + 1)}}^{{\text{th}}}} and (r + 2)th{{\text{(r + 2)}}^{{\text{th}}}} terms in the binomial expansion of (1+x)n{\left( {1 + x} \right)^n} are in A.P., then show that n2−(4r+1)n+4r2−2=0{n^2} - (4r + 1)n + 4{r^2} - 2 = 0 .

Explanation

Solution

Write down the general form of rth{\text{}}{{\text{r}}^{{\text{th}}}} term of binomial expansion of (1+x)n{\left( {1 + x} \right)^n}. For a set of values in arithmetic progression, the sum of the first and third term in the set is equal to twice of the second term.

Formula used:
⇒\Rightarrow The nth{n^{th}} term of the binomial expansion of (1+x)n{(1 + x)^n} is: nCrxr{}^n{C_r}{x^r} .
⇒\Rightarrow For three terms an,an+1,an+2{a_n},{a_{n + 1}},{a_{n + 2}} in arithmetic progression, 2an+1=an+an+22{a_{n + 1}} = {a_n} + {a_{n + 2}} .
⇒\Rightarrow nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}

Complete step by step solution:
We’ve been given that the rth{\text{}}{{\text{r}}^{{\text{th}}}} , (r + 1)th{{\text{(r + 1)}}^{{\text{th}}}} and (r + 2)th{{\text{(r + 2)}}^{{\text{th}}}} terms of the binomial expression of (1+x)n{\left( {1 + x} \right)^n} are in arithmetic progression. Let’s start by finding these terms in the binomial expression. We know that the rth{\text{}}{{\text{r}}^{{\text{th}}}} term of the binomial expression can be written as:
⇒rth\Rightarrow {\text{}}{{\text{r}}^{{\text{th}}}} term: nCr−1(1)n−r+1xr−1=nCr−1xr+1{}^n{C_{r - 1}}{(1)^{n - r + 1}}{x^{r - 1}} = {}^n{C_{r - 1}}{x^{r + 1}}
Similarly, we can write
⇒(r + 1)th\Rightarrow {{\text{(r + 1)}}^{{\text{th}}}} term =nCrxr= {}^n{C_r}{x^r} and
⇒(r + 2)th\Rightarrow {{\text{(r + 2)}}^{{\text{th}}}} term =nCr+1xr+1= {}^n{C_{r + 1}}{x^{r + 1}} .
Now since the coefficients of these three terms are in arithmetic progression, we can write
⇒2nCr=nCr−1+nCr+1\Rightarrow 2{}^n{C_r} = {}^n{C_{r - 1}} + {}^n{C_{r + 1}}
Dividing both sides by nCr{}^n{C_r} , we get
⇒2=nCr−1nCr+nCr+1nCr\Rightarrow 2 = \dfrac{{{}^n{C_{r - 1}}}}{{{}^n{C_r}}} + \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}}
Since nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} , we can write
⇒2=n!(r−1)!(n−(r−1))!n!r!(n−r)!+n!(r+1)!(n−(r+1))!n!r!(n−r)!\Rightarrow 2 = \dfrac{{\dfrac{{n!}}{{(r - 1)!(n - (r - 1))!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}} + \dfrac{{\dfrac{{n!}}{{(r + 1)!(n - (r + 1))!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}}
Using the property that r!=r×(r−1)!r! = r \times (r - 1)! , we can simplify the above expression as
⇒2=rn−r+1+n−rr+1\Rightarrow 2 = \dfrac{r}{{n - r + 1}} + \dfrac{{n - r}}{{r + 1}}
Taking the LCM of the two terms on the right hand side of the equation, we get:
⇒2=r(r+1)+(n−r)(n−r+1)(n−r+1)(r+1)\Rightarrow 2 = \dfrac{{r(r + 1) + (n - r)(n - r + 1)}}{{(n - r + 1)(r + 1)}}
On simplifying the numerator, we get:
⇒2=(r2+r)+(n2−2nr+r2+n−r)(n−r+1)(r+1)\Rightarrow 2 = \dfrac{{({r^2} + r) + ({n^2} - 2nr + {r^2} + n - r)}}{{(n - r + 1)(r + 1)}}
On multiplying both sides by (n−r+1)(r+1)(n - r + 1)(r + 1) , we get
⇒2(n−r+1)(r+1)=(r2+r)+(n2−2nr+r2+n−r)\Rightarrow 2(n - r + 1)(r + 1) = ({r^2} + r) + ({n^2} - 2nr + {r^2} + n - r)
which can be further simplified to
⇒n2−(4r+1)n+4r2−2=0\Rightarrow {n^2} - (4r + 1)n + 4{r^2} - 2 = 0
Hence, we have proved that if the coefficients of rth{\text{}}{{\text{r}}^{{\text{th}}}} , (r + 1)th{{\text{(r + 1)}}^{{\text{th}}}} and (r + 2)th{{\text{(r + 2)}}^{{\text{th}}}} terms in the binomial expansion of (1+x)n{\left( {1 + x} \right)^n} are in A.P. , n2−(4r+1)n+4r2−2=0{n^2} - (4r + 1)n + 4{r^2} - 2 = 0 .

Note:
Since the terms of the binomial expansion are in A.P., we must realize to use the property 2an+1=an+an+22{a_{n + 1}} = {a_n} + {a_{n + 2}} which helps us in relating the coefficients of the binomial expansion. Certain identities of permutation/combination operations can also help in simplifying the solution for e.g. nCr−1nCr=rn−r+1\dfrac{{{}^n{C_{r - 1}}}}{{{}^n{C_r}}} = \dfrac{r}{{n - r + 1}} .