Given-4-2-1-1-2-is-a-geometric-progression-Find-the-sum-of-n-2-terms-of-this-progression-in-terms-of-n-

Question Number 122391 by ZiYangLee last updated on 16/Nov/20

Given 4, 2, 1, \frac{1}{2}, \dots is a geometric progression .

Find the sum of (n + 2) terms of this progression

in terms of n .

Answered by floor(10²Eta[1]) last updated on 16/Nov/20

GP (4, 2, 1, \frac{1}{2}, \dots), q = \frac{1}{2}

\Rightarrow S_{n} = \frac{a_{1} (1 - q^{n})}{1 - q}

S_{n + 2} = \frac{a_{1} (1 - q^{n + 2})}{1 - q} = \frac{4 (1 - {(\frac{1}{2})}^{n + 2})}{1 - \frac{1}{2}}

= \frac{2^{n + 2} - 1}{2^{n - 1}}

Answered by Dwaipayan Shikari last updated on 16/Nov/20

4 + 4 a + 4 a^{2} + 4 a^{3} + \dots + 4 a^{n + 1}

4 (1 + a + a^{2} + a^{3} + \dots + a^{n + 1})

= 4 \frac{1 - a^{n + 2}}{1 - a} (a = \frac{1}{2})

= 8 (1 - \frac{1}{2^{n + 2}}) = 2^{3} - 2^{1 - n}