Question
Question: Find the sum to \[n\] terms of the sequence \[8,88,888,8888,...............\]...
Find the sum to n terms of the sequence 8,88,888,8888,...............
Solution
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 nterms are Sn
i.e. Sn=8+88+888+8888+...........................+n terms
Taking out 8common in all terms we get
Sn=8(1+11+111+1111..............................n terms)
Multiplying and dividing with 9in numerator and denominator we get
Sn=98(9+99+999+9999+..................n terms)
We can rewrite this as
Separating the terms, we get
{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$$terms is equal to $$\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$$ when $$r > 1$$ Since here $$a = 10,{\text{ }}r = 10$$ and sum of $$n$$ one`s is equal to $$n$$.Then the sum of $$n$$ terms is equal to{S_n} = \dfrac{8}{9}\left[ {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{{10 - 1}} - n} \right] \\
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{S_n} = \dfrac{8}{9}\left[ {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right] \\
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{S_n} = \dfrac{{80}}{{81}}\left[ {{{10}^n} - 1} \right] - \dfrac{8}{9}n \\
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