Question

Find the sum to $$n$$ terms of the sequence $$8,88,888,8888,...............$$

Answer 1

Hint: A sequence is an ordered list of numbers. And the dots mean to continue forward in the pattern established in the given sequence. Also, each term in the sequence is called a term.

Step by step solution:
The given sequence is $$8,88,888,8888,...............$$
Let the sum of the given $$n$$terms are $$S_n$$
i.e. $$S_n=8+88+888+8888+...........................+n\text{ terms}$$
Taking out $$8$$common in all terms we get
$$S_n=8\left(1+11+111+1111..............................n\text{ terms}\right)$$
Multiplying and dividing with $$9$$in numerator and denominator we get
$$S_n={\displaystyle \frac{8}{9}}\left(9+99+999+9999+..................n\text{ terms}\right)$$
We can rewrite this as
$$\begin{gathered} S_n={\displaystyle \frac{8}{9}}\left[\left(10-1\right)+\left(100-1\right)+\left(1000-1\right)+.......................................n\text{ terms}\right] \\ {S}_n={\displaystyle \frac{8}{9}}\left[\left(10-1\right)+\left({10}^2-1\right)+\left({10}^3-1\right)+...............................n\text{ terms}\right] \end{gathered}$$
Separating the terms, we get
$$S_n={\displaystyle \frac{8}{9}}\left[\left(10+{10}^2+{10}^3+...............n\text{ terms}\right)-\left(1+1+1+1+...............n\text{ terms}\right)\right]$$
We know that if $$n$$ terms are in G.P. with a common ratio $$r$$and first term $$a$$ then the sum of
the $$n$$terms is equal to $${\displaystyle \frac{a\left(r^n-1\right)}{r-1}}$$ when $$r>1$$

Since here $$a=10,r=10$$ and sum of $$n$$ one`s is equal to $$n$$.Then the sum of $$n$$
terms is equal to
$$\begin{gathered} S_n={\displaystyle \frac{8}{9}}\left[{\displaystyle \frac{10\left({10}^n-1\right)}{10-1}}-n\right] \\ {S}_n={\displaystyle \frac{8}{9}}\left[{\displaystyle \frac{10\left({10}^n-1\right)}{9}}-n\right] \\ {S}_n={\displaystyle \frac{80}{81}}\left[{10}^n-1\right]-{\displaystyle \frac{8}{9}}n \end{gathered}$$
$$\therefore S_n={\displaystyle \frac{80}{81}}\left[{10}^n-1\right]-{\displaystyle \frac{8}{9}}n$$

Therefore, the sum of the terms $$8,88,888,8888,...............$$ is $${\displaystyle \frac{80}{81}}\left[{10}^n-1\right]-{\displaystyle \frac{8}{9}}n$$.

Note: In these types of problems first rewrite the given sequence so the they are in some
progressions like A.P., G.P. or in H.P. By doing this we can sum up them easily by using the known formulae