Question

If $s_n=1+q+q^2+...+q^n$ & $S_n=1+\frac{q+1}{2}+{\left(\frac{q+1}{2}\right)}^2+...+{\left(\frac{q+1}{2}\right)}^n,q\ne 1$, then ${}^{n+1}{C}_1+{}^{n+1}{C}_2.s_1+{}^{n+1}{C}_3.s_2+...+{}^{n+1}{C}_{n+1}.s_n=k^n.S_n$. Find k

Answer 1

$s_n=1+q+q^2+......+q^n=\frac{q^{n+1}-1}{q-1}$

$S_n=1+\left(\frac{q+1}{2}\right)+(\frac{q+1}{2}{)}^2+....+(\frac{q+1}{2}{)}^n=\frac{\left(\frac{(q+1{)}^{n+1}}{2}\right)-1}{\frac{q+1}{2}-1}=\frac{(\frac{q+1}{2}{)}^{n+1}-1}{\frac{q-1}{2}}$

$(n+1)C_1+(n+1)C_2{s}_1+......{+}^{n+1}{C}_{n+1}{S}_n=k^n{S}_n$

LHS ${\Rightarrow }^{n+1}{C}_1{+}^{n+1}{C}_2\frac{[q^2-1]}{(q-1)}{+}^{n+1}{C}_3\frac{[q^3-1]}{(q-1)}+......{+}^{n+1}{C}_{n+1}\frac{[q^{n+1}-1]}{\left[\right(q-1\left)\right]}$

${}^{n+1}{C}_1.\frac{(q-1)}{(q-1)}{+}^{n+1}{C}_2\frac{[q^2-1]}{(q-1)}+.....{+}^{n+1}{C}_{n+1}\frac{[q^{n+1}-1]}{\left(\right(q-1\left)\right)}$

$=\frac{{[}^{(n+1)}{C}_{2}q{+}^{(n+1)}{C}_2{q}^2+....{+}^{(n+1)}{C}_{(n+1)}{q}^{(n+1}\left)\right]{[}^{(n+1)}{C}_1{+}^{(n+1)}{C}_2....{+}^{(n+1)}{C}_3]}{(q-1)}$

$=\frac{\left[\right(1+q{)}^{n+1}-1]-[2^{n+1}-1]}{(q-1)}$

$\frac{2^{n+1}(\frac{1+q}{2}{)}^{n+1}-1}{q-1}=\frac{2^n\left[\right(\frac{1+q}{2}{)}^{n+1}-1]}{\frac{q-1}{2}}$

$=2^{n}Sn.$

$=k^n=Sn$

$\therefore k=2$