Question

If a,b,c,d are any four consecutive coefficients of (1+x) then which of the following is (are) correct

• A. $\frac{a}{a+b}+\frac{c}{c+d}=\frac{2b}{b+c}$
• B. ${\left(\frac{b}{b+c}\right)}^2>\frac{ac}{\left(a+b\right)\left(c+d\right)}$
• C. ${\left(\frac{b}{b+c}\right)}^2<\frac{ac}{\left(a+b\right)\left(c+d\right)}$
• D. ${\left(\frac{b}{b+c}\right)}^2=\frac{ac}{\left(a+b\right)\left(c+d\right)}$

Answer 1

The correct option is A $\frac{a}{a+b}+\frac{c}{c+d}=\frac{2b}{b+c}$

Let $a{}^n{C}_{r-2}$

$b{}^n{C}_{r-1}$

$c{}^n{C}_r$

$d{}^n{C}_{r+1}$

$\frac{a}{a+b}=\frac{{}^n{C}_{r-2}}{{}^n{C}_{r-2}{+}^n{C}_{r-1}}=\frac{{}^n{C}_{r-2}}{{}^{n+1}{C}_{r-2}}=\frac{n!(n-r+3)!(r-2)!}{(n+1)!(r-2)!(n-r+2)}$

$\frac{a}{a+b}=\frac{n-r+3}{n+1}$

$\frac{b}{b+c}=\frac{{}^n{C}_{r-1}}{{}^n{C}_{r-1}{+}^n{C}_r}=\frac{{}^n{C}_{r-1}}{{}^{n+1}{C}_{r-1}}=\frac{n!(n-r+2)!(r-1)!}{(n+1)!(r-1)!(n-r+1)}$

$\frac{b}{b+c}=\frac{n-r+2}{n+1}$

$\frac{c}{c+d}=\frac{{}^n{C}_r}{{}^n{C}_r{+}^n{C}_{r-1}}=\frac{{}^n{C}_r}{{}^{n+1}{C}_r}=\frac{n!(n-r+1)!\left(r\right)!}{(n+1)!\left(r\right)!(n-r)}$

$\frac{c}{c+d}=\frac{n-r+1}{n+1}$

$\frac{a}{a+b}+\frac{c}{c+d}\frac{2n-2r+4}{n+1}=\left(\frac{n-r+2}{n+1}\right)=\frac{2b}{b+c}$

$A$ is correct