Question

If $S_n=1+q+q^2+q^3+...+q^n$ and ${S^{\prime }}_n=1+\left(\frac{q+1}{2}\right)+{\left(\frac{q+1}{2}\right)}^2+...+{\left(\frac{q+1}{2}\right)}^n,q\ne 1$ then ${}^{n+1}{C}_1{+}^{n+1}{C}_2.S_1{+}^{n+1}{C}_3.S_2+...{+}^{n+1}{C}_{n+1}.S_n=$

• A. $2^{n-1}.{S^{\prime }}_n$
• B. $2^n.{S^{\prime }}_n$
• C. $2^{n+1}.{S^{\prime }}_n$
• D. None of these

Answer 1

Answer: (B)

$2^n.{S^{\prime }}_n$

$S_n$=$\frac{q^{n+1}-1}{q-1}$ summation of g.p

${}^{n+1}{C}_1$+${}^{n+1}{C}_2$$\frac{q^2-1}{q-1}$+${}^{n+1}{C}_3$$\frac{q^3-1}{q-1}$+..................+${}^{n+1}{C}_{n+1}$$\frac{q^{n+1}-1}{q-1}$

$\frac{{}^{n+1}{C}_0}{q-1}$-${}^{n+1}{C}_1$$\frac{q-1}{q-1}$+${}^{n+1}{C}_2$$\frac{q^2-1}{q-1}$+${}^{n+1}{C}_3$$\frac{q^3-1}{q-1}$+..................+${}^{n+1}{C}_{n+1}$$\frac{q^{n+1}-1}{q-1}$-$\frac{{}^{n+1}{C}_0}{n-1}$

($\frac{{}^{n+1}{C}_0}{q-1}$+${}^{n+1}{C}_1$$\frac{q}{q-1}$+${}^{n+1}{C}_2$$\frac{q^2}{q-1}$+${}^{n+1}{C}_3$$\frac{q^3}{q-1}$+..................+${}^{n+1}{C}_{n+1}$$\frac{q^{n+1}}{q-1}$)-($\frac{{}^{n+1}{C}_0}{q-1}$+${}^{n+1}{C}_1$$\frac{}{q-1}$+${}^{n+1}{C}_2$$\frac{}{q-1}$+${}^{n+1}{C}_3$$\frac{}{q-1}$+.................. +${}^{n+1}{C}_{n+1}$$\frac{}{q-1}$)

=$\frac{{(1+q)}^{n+1}}{q-1}$-$\frac{2^{n+1}}{q-1}$

=$2^n(\frac{(\frac{1+q}{2}{)}^{n+1}}{\frac{q-1}{2}}$-$\frac{2}{q-1})$

=$2^n(\frac{(\frac{1+q}{2}{)}^{n+1}}{\frac{q-1}{2}}$-$\frac{1}{\frac{q-1}{2}})$

=$2^n$$S_n^{\prime }$

and$S_n^{\prime }$=$(\frac{(\frac{1+q}{2}{)}^{n+1}}{\frac{q-1}{2}}$$-$$\frac{1}{\frac{q-1}{2}})$ summation of g.p