Question

# Find the sum of the following series up to $$n$$ terms: $$.6+.66+.666+....$$

Solution

## Let,  $$\displaystyle \quad { S }_{ n }=.6+.66+.666+...\quad to\quad n\quad terms\quad \\ =6\left[ 0.1+0.11+0.111+...\quad to\quad n\quad terms \right] \\ =\dfrac { 6 }{ 9 } \left[ 0.9+0.99+0.999+...\quad to\quad n\quad terms \right] \\\\ =\dfrac { 6 }{ 9 } \left[ \left( 1-\dfrac { 1 }{ 10 } \right) +\left( 1-\dfrac { 1 }{ { 10 }^{ 2 } } \right) +\left( 1-\dfrac { 1 }{ { 10 }^{ 3 } } \right) +...to\quad n\quad terms \right] \\\\ =\dfrac { 2 }{ 3 } \left[ (1+{ 1 }+...\quad n\quad terms)-\dfrac { 1 }{ 10 } \left( 1+\dfrac { 1 }{ 10 } +\dfrac { 1 }{ { 10 }^{ 2 } } +...\quad n\quad terms \right) \right] \\ =\dfrac { 2 }{ 3 } \left[ n-\dfrac { 1 }{ 10 } \left( \dfrac { 1-\left( \dfrac { 1 }{ 10 } \right) ^{ n } }{ 1-\dfrac { 1 }{ 10 } } \right) \right] \\ =\dfrac { 2 }{ 3 } n-\dfrac { 2 }{ 30 } \times \dfrac { 10 }{ 9 } (1-{ 10 }^{ -n })\\ =\dfrac { 2 }{ 3 } n-\dfrac { 2 }{ 27 } (1-{ 10 }^{ -n })$$MathematicsNCERTStandard XI

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