Question

The sum of n terms of the series $1.4+3.04+5.004+7.0004+..$ is

• A. $n^2+\frac{4}{9}(1+\frac{1}{{10}^n})$
• B. $n^2+\frac{4}{9}(1-\frac{1}{{10}^n})$
• C. $n+\frac{4}{9}(1-\frac{1}{{10}^n})$
• D. none of these

Answer 1

Answer: (B)

$n^2+\frac{4}{9}(1-\frac{1}{{10}^n})$

the series $1.4+3.04+5.004+7.0004+..$

can be written as $1+0.4+3+0.04+5+0.004+7+0.0004+...$

rearranged as an A.P $1+3+5+7+...$ + an G.P $0.4+0.04+0.004+0.0004+..$

So in A.P $a=1,d=2$

So in G.P $a=0.4,r=0.1$

Sum of n terms is $\frac{n}{2}(2a+(n-1\left)d\right)$ of A.P + $\frac{a(r^n-1)}{r-1}$ of G.P

Sum of n terms is $\frac{n}{2}(2+(n-1)2+\frac{0.4(1-(0.1{)}^n)}{1-0.1}$

Sum of n terms is $\frac{n}{2}2n+\frac{0.4(1-\frac{1}{{10}^n})}{0.9}$

Sum of n terms is $n^2+\frac{4}{9}(1-\frac{1}{{10}^n})$