\(\displaystyle\sum_{k=1}^{11} (2+3^{k})\)
\(\displaystyle\sum_{k=1}^{11} (2)+ \displaystyle\sum_{k=1}^{11} (3^{k})= 2(11)+\displaystyle\sum_{k=1}^{11} (3^{k})=22+\displaystyle\sum_{k=1}^{11} (3^{k}) ....(1) \displaystyle\sum_{k=1}^{11} 3^{k}= 3^1+3^2+3^3+...3^{11}\)
The terms of this sequence 3, \(3^2,3^3\) , … forms a G.P.
\(s_{n} = \frac{a(r^n-1)}{r-1}\)
⇒ \(s_{11}\) = \(\frac{3[(3)^{11}-1]}{3-1}\)
⇒ \(s_{11}=\frac{3}{2}3^{11}-1\)
∴\(\displaystyle\sum_{k=1}^{11} 3^{k}=\frac{3}{2}(3^{11}-1)\)
Substituting this value in equation (1), we obtain
\(\displaystyle\sum_{k=1}^{11} (2+3^{k}) = 22+\frac{3}{2}(3^{11}-1)\)
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab) n .