Question

Find the sum to n terms of the series 9+99+999

Answer 1

Refer to the attachment below.
Hope it helps

Answer 2

Answer:

$Sum=\frac{10^{n+1}-9n-10}{9}$

Step-by-step explanation:

In the question,

We have to find the sum of 'n' terms of the series,

9 + 99 + 999 + 9999 + .............

So,

For finding the sum of the given series, we can write the series as,

= (10 - 1) + (100 - 1) + (1000 - 1) + ..............

= (10 - 1) + (10² - 1) + (10³ - 1) + ..............

= (10 + 10² + 10³ + ........ + 10ⁿ) - (1 + 1 + 1 + ..............n terms)

From the Sum of the GP, we can say that,

$Sum=\frac{10(10^{n}-1)}{10 - 1}=\frac{10(10^{n}-1)}{9}$

Also,

Sum of 'n' number of 1 is = n

Now, the Sum of the given series will be, therefore,

$Sum=\frac{10(10^{n}-1)}{9}-n\\Sum=\frac{10^{n+1}-10-9n}{9}\\Sum=\frac{10^{n+1}-9n-10}{9}$

Therefore, the required sum of the series is,

$Sum=\frac{10^{n+1}-9n-10}{9}$