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Question: The sum of the series \(3 + 33 + 333 + \ldots + n\) terms is....

The sum of the series 3+33+333+…+n3 + 33 + 333 + \ldots + n terms is.

A

127(10n+1+9n−28)\frac { 1 } { 27 } \left( 10 ^ { n + 1 } + 9 n - 28 \right)

B

127(10n+1−9n−10)\frac { 1 } { 27 } \left( 10 ^ { n + 1 } - 9 n - 10 \right)

C

127(10n+1+10n−9)\frac { 1 } { 27 } \left( 10 ^ { n + 1 } + 10 n - 9 \right)

D

None of these

Answer

127(10n+1−9n−10)\frac { 1 } { 27 } \left( 10 ^ { n + 1 } - 9 n - 10 \right)

Explanation

Solution

Series 3 + 33 + 333 +…......+ n terms

Given series can be written as,

=13[9+99+999+………+n= \frac { 1 } { 3 } [ 9 + 99 + 999 + \ldots \ldots \ldots + n terms ]]

=13[(10−1)+(102−1)+(103−1)+….+n= \frac { 1 } { 3 } \left[ ( 10 - 1 ) + \left( 10 ^ { 2 } - 1 \right) + \left( 10 ^ { 3 } - 1 \right) + \ldots . + n \right. terms ]]

=13[10+102+….+10n]= \frac { 1 } { 3 } \left[ 10 + 10 ^ { 2 } + \ldots . + 10 ^ { n } \right] −13[1+1+1+….+n- \frac { 1 } { 3 } [ 1 + 1 + 1 + \ldots . + n terms ]]

=13⋅10(10n−1)10−1−13⋅n= \frac { 1 } { 3 } \cdot \frac { 10 \left( 10 ^ { n } - 1 \right) } { 10 - 1 } - \frac { 1 } { 3 } \cdot n =13[10n+1−109−n]= \frac { 1 } { 3 } \left[ \frac { 10 ^ { n + 1 } - 10 } { 9 } - n \right]

=13[10n+1−9n−109]= \frac { 1 } { 3 } \left[ \frac { 10 ^ { n + 1 } - 9 n - 10 } { 9 } \right] =127[10n+1−9n−10]= \frac { 1 } { 27 } \left[ 10 ^ { n + 1 } - 9 n - 10 \right] .