Solveeit Logo
Login

Question

Mathematics Question on Sequence and series

If SmS_m denotes the sum of first mm terms of a G.P.G.P. of nn terms with common ratio rr, then the sum of their products taken two by two is

A

SmSm−1S_mS_{m-1}

B

rr+1SmSm−1\frac{r}{r+1}S_mS_{m-1}

C

rr−1SmSm−1\frac{r}{r-1}S_mS_{m-1}

D

r+1rSmSm−1\frac{r+1}{r}S_mS_{m-1}

Answer

rr+1SmSm−1\frac{r}{r+1}S_mS_{m-1}

Explanation

Solution

Sm=a[1−rm]1−rS_{m} = \frac{a\left[1-r^{m}\right]}{1-r} Now\left({\text{sum ofmterms}}\right)^{2} = \left({\text{sum of squares ofmterms}}\right) + 2T where TT denotes the sum of the products taken two by two. ∴2T=[a(1−rm)1−r]2−a2[1−r2m1−r2]\therefore 2T = \left[\frac{a\left(1-r^{m}\right)}{1-r}\right]^{2} -a^{2} \left[\frac{ 1-r^{2}m}{1-r^{2}}\right] =a2(1−rm)2(1−r)2−a2(1−rm)(1+rm)(1−r)(1+r) = a^{2} \frac{\left(1-r^{m}\right)^{2}}{\left(1-r\right)^{2}}-a^{2} \frac{\left(1-r^{m}\right)\left(1+r^{m}\right)}{\left(1-r\right)\left(1+r\right)} =a2(1−rm)(1−r)[1−rm1−r−1+rm1+r] = a^{2 } \frac{\left(1-r^{m}\right)}{\left(1-r\right)}\left[\frac{1-r^{m}}{1-r} - \frac{1+r^{m}}{1+r}\right] =a2(1−rm)(1−r)= a^{2} \frac{\left(1-r^{m}\right)}{\left(1-r\right)} =[(1−rm+r−rm+1)−(1+rm−r−rm+1)](1−r)(1+r)= \frac{\left[ \left(1-r^{m}+r -r^{m+1}\right) - \left(1+r^{m}-r-r^{m+1}\right)\right]}{\left(1-r\right)\left(1+r\right)} ∴2T=a2(1−rm)1−r2r(1−rm−1)(1−r)(1+r)\therefore 2T = \frac{a^{2}\left(1-r^{m}\right)}{1-r} \frac{2r\left(1-r^{m-1}\right)}{\left(1-r\right)\left(1+r\right)} ∴T=a(1−rm)1−ra(1−rm−1−1)1−rr1+r\therefore T = \frac{a\left(1-r^{m}\right)}{1-r} \frac{a\left(1-r^{m-1}-1\right)}{1-r} \frac{r}{1+r} =Sm⋅Sm−1⋅rr+1= S_{m}\cdot S_{m-1} \cdot \frac{r}{r+1} Hence T=rr+1Sm⋅Sm−1T= \frac{r}{r+1} S_{m} \cdot S_{m-1}