Question

Find sum of n term:8,88,888,_ _ _ _ _ _

Answer 1

The given sequence is: 8, 88, 888, ...

We can observe a pattern here. Each term is comprised of the digit "8" repeated a certain number of times:

- The 1st term is \( 8 \) (one '8').
- The 2nd term is \( 88 \) (two '8's).
- The 3rd term is \( 888 \) (three '8's).

In general, the \( n \)-th term of the sequence can be expressed as:

\[
a_n = \underbrace{88...8}_{n \text{ times}}
\]

This can also be mathematically represented as:

\[
a_n = 8 \times \frac{10^n - 1}{9}
\]

This formula arises because the \( n \)-digit number made up entirely of '8's can be expressed as \( 8 \times 10^{n-1} + 8 \times 10^{n-2} + ... + 8 \), which simplifies to \( 8 \times \frac{10^n - 1}{9} \).

Now, we want to find the sum of the first \( n \) terms, \( S_n \):

\[
S_n = a_1 + a_2 + a_3 + ... + a_n
\]

Substituting in our formula for \( a_n \):

\[
S_n = 8 \times \frac{10^1 - 1}{9} + 8 \times \frac{10^2 - 1}{9} + 8 \times \frac{10^3 - 1}{9} + ... + 8 \times \frac{10^n - 1}{9}
\]

Factoring out the \( 8/9 \):

\[
S_n = \frac{8}{9} \left[ (10^1 - 1) + (10^2 - 1) + (10^3 - 1) + ... + (10^n - 1) \right]
\]

\[
S_n = \frac{8}{9} \left[ (10^1 + 10^2 + 10^3 + ... + 10^n) - n \right]
\]

The sum of a geometric series \( 10^1 + 10^2 + ... + 10^n \) can be calculated using the formula:

\[
\sum_{k=1}^n r^k = r \frac{r^n - 1}{r - 1}
\]

In our case, \( r = 10 \):

\[
10^1 + 10^2 + ... + 10^n = 10 \frac{10^n - 1}{10 - 1} = \frac{10(10^n - 1)}{9}
\]

Now substituting this back into our expression for \( S_n \):

\[
S_n = \frac{8}{9} \left[ \frac{10(10^n - 1)}{9} - n \right]
\]

\[
S_n = \frac{8}{9} \cdot \frac{10(10^n - 1)}{9} - \frac{8n}{9}
\]

\[
S_n = \frac{80(10^n - 1)}{81} - \frac{8n}{9}
\]

Thus, the sum of the first \( n \) terms of the sequence is:

\[
S_n = \frac{80(10^n - 1)}{81} - \frac{8n}{9}
\]

This is the final expression for the sum of the first \( n \) terms in the sequence \( 8, 88, 888, \ldots \).