Question

please tell ???
it is from sequence and series​

Answer 1

Given,

$S_1=\dfrac {a(r^n-1)}{r-1}\\\\S_2=\dfrac {a(r^{2n}-1)}{r-1}\\\\S_3=\dfrac {a(r^{3n}-1)}{r-1}$

So,

$S_2-S_1=\dfrac {a(r^{2n}-1)}{r-1}-\dfrac {a(r^n-1)}{r-1}\\\\S_2-S_1=\dfrac {a(r^{2n}-1-r^n+1)}{r-1}\\\\S_2-S_1=\dfrac {a(r^{2n}-r^n)}{r-1}\\\\S_2-S_1=\dfrac {ar^n(r^n-1)}{r-1}$

And,

$S_3-S_2=\dfrac {a(r^{3n}-1)}{r-1}-\dfrac {a(r^{2n}-1)}{r-1}\\\\S_3-S_2=\dfrac {a(r^{3n}-1-r^{2n}+1)}{r-1}\\\\S_3-S_2=\dfrac {a(r^{3n}-r^{2n})}{r-1}\\\\S_3-S_2=\dfrac {ar^{2n}(r^n-1)}{r-1}$

Now,

$\begin {aligned}&\ \ S_1(S_3-S_2)\\\\=\ \ &\dfrac {a(r^n-1)}{r-1}\cdot\dfrac {ar^{2n}(r^n-1)}{r-1}\\\\=\ \ &\left (\dfrac {a(r^n-1)}{r-1}\right)^2\cdot(r^n)^2\\\\=\ \ &\left (\dfrac {ar^n(r^n-1)}{r-1}\right)^2\\\\=\ \ &(S_2-S_1)^2\end {aligned}$

Hence Proved!