Find the sum to $n$ terms of the series $0.7+7.7+0.77+77.7+0.777+777.7+0.7777+....$ where $n$ is even.
- $\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { n }{ 4 } -\displaystyle \frac { \left( 1-\left( 0.1 \right) ^{ n/4 } \right) }{ 9 } \right\} +\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { 10((10)^{ n/4 }-1) }{ 9 } -0.1\displaystyle \frac { n }{ 4 } \right\} $
- $\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { n }{ 2 } -\displaystyle \frac { \left( 1-\left( 0.1\right) ^{ n/2 } \right) }{ 9 } \right\} +\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { 10((10)^{ n/2 }-1) }{ 9 } -\displaystyle \frac { n }{ 2 } \right\} $
- $\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { n }{ 2 } +\displaystyle \frac { \left( 1-\left( 0.1 \right) ^{ n/2 } \right) }{ 9 } \right\} +\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { 10((10)^{ n/2 }-1) }{ 9 } +\displaystyle \frac { n }{ 2 } \right\} $
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none of these
The series is (0.7 + 0.77 + ... n/2 terms) + (7.7 + 77.7 + ... n/2 terms). This is a complex summation problem involving geometric series components. Option B correctly represents the sum of these two distinct parts.
By grouping the odd and even positioned terms, we separate the series into the sum of repeating decimals and multiples of ten, substituting m = n/2 to represent the number of terms in each group. The repeating decimal terms form a progression evaluated using the geometric series sum formula, leading to the summed block (7/9)((n/2) - (1 - 0.1^(n/2))/9). The multiples of ten form a geometric progression evaluated with the standard sum formula, resulting in the second summed block (7/9)((10(10^(n/2) - 1))/9 - n/2). Adding these two blocks yields the total sum to n terms as (7/9)((n/2) - (1 - (0.1)^(n/2))/9) + (7/9)((10((10)^(n/2) - 1))/9 - n/2).