The sum of the first n terms in the sequence 8, 88, 888, 8888, ….. is _____.

The correct option is:
$\frac{80}{81}({10}^n-1)-\frac{8}{9}n$

Step-by-Step Explanation:

We want to find the sum of the first $n$ terms of the sequence:
$S_n=8+88+888+8888+\cdots \text{ to }n\text{ terms}$

As detailed in the step-by-step derivation visible in the provided image, we can find the sum using the following steps:

Step 1: Take 8 as a common factor.
By factoring out 8 from each term, we obtain:
$S_n=8(1+11+111+1111+\cdots \text{ to }n\text{ terms})$

Step 2: Multiply and divide by 9.
To rewrite the terms in a form related to powers of 10, we multiply and divide the expression by 9:
$S_n=\frac{8}{9}[9+99+999+9999+\cdots \text{ to }n\text{ terms}]$

Step 3: Express terms in powers of 10.
Each number consisting of 9s can be written as a power of 10 minus 1 (i.e., $9=10-1$, $99={10}^2-1$, etc.):
$S_n=\frac{8}{9}\left[\right(10-1)+({10}^2-1)+({10}^3-1)+\cdots +({10}^n-1\left)\right]$

Step 4: Group the terms.
Separate the powers of 10 and the constant terms:
$S_n=\frac{8}{9}\left[\right(10+{10}^2+{10}^c+\cdots +{10}^n)-(1+1+1+\cdots \text{ to }n\text{ terms}\left)\right]$

Step 5: Apply the sum formulas.
The first part is a Geometric Progression (GP) with the first term $a=10$, common ratio $r=10$, and $n$ terms. Using the GP sum formula $S_{\text{GP}}=\frac{a(r^n-1)}{r-1}$:
$10+{10}^2+\cdots +{10}^n=\frac{10({10}^n-1)}{10-1}=\frac{10({10}^n-1)}{9}$

The second part is the sum of $n$ terms of 1:
$(1+1+1+\cdots \text{ to }n\text{ terms})=n$

Substitute these values back into the equation:
$S_n=\frac{8}{9}[\frac{10({10}^n-1)}{9}-n]$

Step 6: Simplify the expression.
Distribute $\frac{8}{9}$ across the terms in the bracket:
$S_n=\frac{80}{81}({10}^n-1)-\frac{8}{9}n$

General Trick/Shortcut formula:
For any sequence of the form $N,NN,NNN,\cdots$, the sum of the first $n$ terms is given by the formula:
$S_n=\frac{N}{9}[\frac{10({10}^n-1)}{9}-n]$
Substituting $N=8$ directly yields the correct answer.