Sequence And Series Question 190
Question: The sum of the series $ 3+33+333+…+n $ terms is
[RPET 2000]
Options:
A) $ \frac{1}{27}({10^{n+1}}+9n-28) $
B) $ \frac{1}{27}({10^{n+1}}-9n-10) $
C) $ \frac{1}{27}({10^{n+1}}+10n-9) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Series 3 + 33 + 333 +?……+ n terms Given series can be written as, $ =\frac{1}{3}[9+99+999+……..+n\text{terms }] $ $ =\frac{1}{3}[ (10-1)+(10^{2}-1)+(10^{3}-1)+….+n,terms ] $ $ =\frac{1}{3}[ 10+10^{2}+….+10^{n} ] $ $ -\frac{1}{3}[ 1+1+1+….+n,terms ] $ $ =\frac{1}{3},.,\frac{10,(10^{n}-1)}{10-1}-\frac{1}{3}.n, $ $ =\frac{1}{3}[ \frac{{10^{n+1}}-10}{9}-n ] $ $ =\frac{1}{3},[ \frac{{10^{n,+,1}}-9n-10}{9} ] $ $ =\frac{1}{27}[{10^{n,+,1}}-9n-10] $ .
BETA
