Question

If $C_0,C_1,C_2,\dots ,C_n$ denote the binomial coefficients of the expansion $(1+x{)}^n$ and $\sum _{r=0}^n(-1{)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\dots \text{upto}m\text{terms}]$$=\frac{a^{mn}-1}{b^{mn}(c^n-1)},$ then

• A. $a=2$
• B. $c=2$
• C. $a-b+c=2$
• D. $a+b-c=4$

Answer 1

Answer: (A), (B), (C)

$a-b+c=2$

Let $S=\sum _{r=0}^n(-1{)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\dots \text{upto}m\text{terms}]$
$=\sum _{r=0}^n(-1{)}^r{}^n{C}_r[{\left(\frac{1}{2}\right)}^r+{\left(\frac{3}{4}\right)}^r+{\left(\frac{7}{8}\right)}^r+\dots +{\left(\frac{2^m-1}{2^m}\right)}^r]=\sum _{r=0}^n{}^n{C}_r{(-\frac{1}{2})}^r+\sum _{r=0}^n{}^n{C}_r{(-\frac{3}{4})}^r+\dots +\sum _{r=0}^n{}^n{C}_r{(-\frac{2^m-1}{2^m})}^r={(1-\frac{1}{2})}^n+{(1-\frac{3}{4})}^n+\dots +{(1-\frac{2^m-1}{2^m})}^n=\frac{1}{2^n}+{\left(\frac{1}{2^n}\right)}^2+{\left(\frac{1}{2^n}\right)}^3+\dots +{\left(\frac{1}{2^n}\right)}^m=\frac{\frac{1}{2^n}({\left(\frac{1}{2^n}\right)}^m-1)}{\frac{1}{2^n}-1}=\frac{1-2^{mn}}{2^{mn}(1-2^n)}\therefore S=\frac{2^{mn}-1}{2^{mn}(2^n-1)}$

$\therefore a=2,b=2,c=2$