Australia
What is the sum of the first ^@ 5 ^@ terms of the geometric series ^@ 1, \dfrac{ 2 }{ 3 }, \dfrac{ 4 }{ 9 } \space ...? ^@
Answer:
^@ \dfrac{ 211 }{ 81 } ^@
- The sum of first ^@ n ^@ terms of a ^@ G.P. ^@ is given by,
^@ S_n = \dfrac{ a(1 - r^n) }{ (1 - r) } ^@
Here, the first term, ^@a = 1^@ and
the common ratio, ^@ r = \dfrac{ a_{k+1} }{ a_k } \text{ where } k ≥ 1
^@
^@ \implies r = \dfrac{ \dfrac{ 2 }{ 3 } } { 1 } = \dfrac{ 2 }{ 3 } ^@ - The sum of first ^@ n ^@ terms of this ^@ G.P. ^@ is given by,
@^ \begin{align} S_n &= \dfrac { (1)\left(1 - \left( \dfrac{ 2 }{ 3 } \right)^n \right) } { 1 - \dfrac{ 2 }{ 3 } } \\ &= \dfrac { \left(1 - \left( \dfrac{ 2 }{ 3 } \right)^n \right) } { \dfrac{ 1 }{ 3 } } \\ &= 3 \left[ 1 - \left( \dfrac{ 2 }{ 3 } \right)^n \right] \\ \end{align} @^ Now, the sum of the first ^@ 5 ^@ terms of the ^@G.P^@ is given by, @^ \begin{align} S_5 &= 3 \left[ 1 - \left( \dfrac{ 2 }{ 3 } \right)^5 \right] \\ &= 3 \left[ 1 - \dfrac{ 32 }{ 243 } \right] \\ &= 3 \times \dfrac{ 211 }{ 243 } \\ &= \dfrac{ 211 }{ 81 } \end{align} @^ - Hence, the sum of the first ^@ 5 ^@ terms of the ^@G.P.^@ is ^@ \dfrac{ 211 }{ 81 } ^@.
