The sum of n terms of the series $1.4 + 3.04 + 5.004 + 7.0004+..$ is
- $n^2+\dfrac{4}{9}(1+\dfrac{1}{10^n})$
- $n^2+\dfrac{4}{9}(1-\dfrac{1}{10^n})$
- $n+\dfrac{4}{9}(1-\dfrac{1}{10^n})$
-
none of these
The series is (1+0.4) + (3+0.04) + (5+0.004) + ... = (1+3+5+...+(2n-1)) + (0.4+0.04+0.004+...). The sum of the first part is n^2. The second part is a geometric series: 4/10 + 4/100 + ... = (4/10) * (1 - (1/10)^n) / (1 - 1/10) = (4/10) * (1 - 1/10^n) / (9/10) = 4/9 * (1 - 1/10^n).
The nth term of the series can be split into an arithmetic progression and a geometric progression, written as 2 times n minus 1 plus 4 divided by 10 to the power of n. The sum of the arithmetic part across n terms is n squared. The sum of the geometric part is the series 0.4 plus 0.04 plus 0.004, continuing for n terms, which adds to 4 divided by 9 multiplied by the quantity 1 minus 1 divided by 10 to the power of n. Adding these two sums gives the result of n squared plus 4 divided by 9 multiplied by the quantity 1 minus 1 divided by 10 to the power of n.