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Question: If a, b, c be the three consecutive coefficients in the expansion of a power of \({{\left( 1+x \righ...

If a, b, c be the three consecutive coefficients in the expansion of a power of (1+x)n{{\left( 1+x \right)}^{n}} , then n=?n=?
A. 2ac+b(a+c)b2−ac B. a+b-2c C. 2acb2−ac D. None of these. \begin{aligned} & \text{A}\text{. }\dfrac{2ac+b\left( a+c \right)}{{{b}^{2}}-ac} \\\ & \text{B}\text{. a+b-2c} \\\ & \text{C}\text{. }\dfrac{2ac}{{{b}^{2}}-ac} \\\ & \text{D}\text{. None of these}\text{.} \\\ \end{aligned}

Explanation

Solution

Hint: To find this question first we need to consider a=nCr,b=nCr+1,c=nCr+2a={}^{n}{{C}_{r}},b={}^{n}{{C}_{r+1}},c={}^{n}{{C}_{r+2}} by assuming r,r+1&r+2r,r+1\And r+2 as the three consecutive coefficients. Then find the value of ‘r’ by dividing ab\dfrac{a}{b} & bc\dfrac{b}{c} and equalize them to find the value of ‘n’.

Complete step by step solution:
Let us consider the consecutive coefficients to be r,r+1,r+2r,r+1,r+2 terms.
So, we will consider a, b, c as –
a=nCr b=nCr+1 c=nCr+2 \begin{aligned} & a={}^{n}{{C}_{r}} \\\ & b={}^{n}{{C}_{r+1}} \\\ & c={}^{n}{{C}_{r+2}} \\\ \end{aligned}
Here, we will find the value of ‘r’ by dividing a by b we get –
ab=nCrnCr+1\dfrac{a}{b}=\dfrac{{}^{n}{{C}_{r}}}{{}^{n}{{C}_{r+1}}}
We know that nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} . So, we get –
ab=n!r!(n−r)!n!(r+1)!(n−(r+1))!\dfrac{a}{b}=\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}}
We can also write it as –
ab=n!r!(n−r)!×(r+1)!(n−(r+1))!n!\dfrac{a}{b}=\dfrac{n!}{r!\left( n-r \right)!}\times \dfrac{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}{n!}
By simplifying this we get –
ab=n!r!(n−r)!×(r+1)!(n−r−1)!n!\dfrac{a}{b}=\dfrac{n!}{r!\left( n-r \right)!}\times \dfrac{\left( r+1 \right)!\left( n-r-1 \right)!}{n!}
By cancelling the common factors from numerator and denominator, we get –
ab=(r+1)!(n−r−1)!r!(n−r)!\dfrac{a}{b}=\dfrac{\left( r+1 \right)!\left( n-r-1 \right)!}{r!\left( n-r \right)!}
Here, (r+1)!\left( r+1 \right)! can be written as (r+1)r!\left( r+1 \right)r! and (n−r)!\left( n-r \right)! can be written as (n−r)(n−r−1)!\left( n-r \right)\left( n-r-1 \right)!
So we get –
ab=(r+1)r!(n−r−1)!r!(n−r)(n−r−1)!\dfrac{a}{b}=\dfrac{\left( r+1 \right)r!\left( n-r-1 \right)!}{r!\left( n-r \right)\left( n-r-1 \right)!}
By cancelling the common factors from numerator and denominator, we get –
ab=r+1n−r\dfrac{a}{b}=\dfrac{r+1}{n-r}
By cross multiplication, we get –
a(n−r)=b(r+1) an−ar=br+b \begin{aligned} & a\left( n-r \right)=b\left( r+1 \right) \\\ & an-ar=br+b \\\ \end{aligned}
By subtracting ‘b’ and adding ‘ar’ on both sides, we get –
an−b=br+ar an−b=r(a+b) \begin{aligned} & an-b=br+ar \\\ & an-b=r\left( a+b \right) \\\ \end{aligned}
By dividing a+ba+b on the both sides, we get –
r=an−ba+br=\dfrac{an-b}{a+b} …………………….. (1)
Now we will divide b and c so we get an another value of ‘r’, so we get –
bc=nCr+1nCr+2\dfrac{b}{c}=\dfrac{{}^{n}{{C}_{r+1}}}{{}^{n}{{C}_{r+2}}}
We know that nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} . So, we get –
bc=n!(r+1)!(n−(r+1))!n!(r+2)!(n−(r+2))!\dfrac{b}{c}=\dfrac{\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}}{\dfrac{n!}{\left( r+2 \right)!\left( n-\left( r+2 \right) \right)!}}
We can also write it as –
bc=n!(r+1)!(n−(r+1))!×(r+2)!(n−(r+2))!n!\dfrac{b}{c}=\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}\times \dfrac{\left( r+2 \right)!\left( n-\left( r+2 \right) \right)!}{n!}
By simplifying this we get –
bc=n!(r+1)!(n−r−1)!×(r+2)!(n−r−2)!n!\dfrac{b}{c}=\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}\times \dfrac{\left( r+2 \right)!\left( n-r-2 \right)!}{n!}
By cancelling the common factors from numerator and denominator, we get –
bc=(r+2)!(n−r−2)!(r+1)!(n−r−1)!\dfrac{b}{c}=\dfrac{\left( r+2 \right)!\left( n-r-2 \right)!}{\left( r+1 \right)!\left( n-r-1 \right)!}
Here, (r+2)!\left( r+2 \right)! can be written as (r+2)(r+1)!\left( r+2 \right)\left( r+1 \right)! and (n−r−1)!\left( n-r-1 \right)! can be written as (n−r−1)(n−r−2)!\left( n-r-1 \right)\left( n-r-2 \right)!
So we get –
ab=(r+2)(r+1)!(n−r−2)!(r+1)!(n−r−1)(n−r−2)!\dfrac{a}{b}=\dfrac{\left( r+2 \right)\left( r+1 \right)!\left( n-r-2 \right)!}{\left( r+1 \right)!\left( n-r-1 \right)\left( n-r-2 \right)!}
By cancelling the common factors from numerator and denominator, we get –
bc=r+2n−r−1\dfrac{b}{c}=\dfrac{r+2}{n-r-1}
By cross multiplication, we get –
b(n−r−1)=c(r+2) bn−br−b=cr+2c \begin{aligned} & b\left( n-r-1 \right)=c\left( r+2 \right) \\\ & bn-br-b=cr+2c \\\ \end{aligned}
By subtracting ‘2c2c ’ and adding ‘brbr’ on both sides, we get –
bn−b−2c=cr+br bn−b−2c=r(b+c) \begin{aligned} & bn-b-2c=cr+br \\\ & bn-b-2c=r\left( b+c \right) \\\ \end{aligned}
By dividing b+cb+c on the both sides, we get –
r=bn−b−2cb+cr=\dfrac{bn-b-2c}{b+c} …………………….. (2)
By equalizing the values of ‘r’ from equation (1) and (2) to get the value of ‘n’, we get –
an−ba+b=bn−b−2cb+c\dfrac{an-b}{a+b}=\dfrac{bn-b-2c}{b+c}
By cross multiplication we get –
(an−b)(b+c)=(bn−b−2c)(a+b)\left( an-b \right)\left( b+c \right)=\left( bn-b-2c \right)\left( a+b \right)
By simplifying we get –
abn+acn−b2−bc=abn+b2n−ab−b2−2ac−2bcabn+acn-{{b}^{2}}-bc=abn+{{b}^{2}}n-ab-{{b}^{2}}-2ac-2bc
By cancelling and transferring the above equation, we get –
ab+2ac+2bc−bc=−acn+b2nab+2ac+2bc-bc=-acn+{{b}^{2}}n
By taking ‘n’ as common at one side, we get –
ab+2a+bc=n(b2−ac)ab+2a+bc=n\left( {{b}^{2}}-ac \right)
By dividing ac−b2ac-{{b}^{2}}on both sides, we get –
n=bc+ab+2acb2−acn=\dfrac{bc+ab+2ac}{{{b}^{2}}-ac}
Hence, option (A) is the correct answer.

Note: Generally students can make mistakes while solving this question. They may take x,x2,x3x,{{x}^{2}},{{x}^{3}} as the three consecutive coefficient terms or can take directly the consecutive coefficients (r,r+1,r+2)\left( r,r+1,r+2 \right) as a, b, c, which may lead to get them a wrong answer.