Question

S₁, S₂, S₃, ..... , S_n are the sums of n infinite geometric progressions. The first term S of these progressions are 1, 2² – 1, 2³ – 1, 2⁴ – 1, ..... , 2ⁿ – 1 and the common ratios are \(\frac{1}{2},\frac{1}{2^2},\frac{1}{2^3},.....\frac{1}{2^n}.\) Calculate the sum S₁ + S₂ + S₃ + ..... + S_n.

Answer 1

Since S_∞ = \(\frac{a}{1-r},\) 

S₁ = \(\frac{1}{1-\frac{1}{2}}\) = \(\frac{1}{\frac{1}{2}}\) = 2

S₂ = \(\frac{2^2-1}{1-\frac{1}{2^2}}\) = \(\frac{2^2-1}{\frac{2^2-1}{2^2}}\) = 2²

S₃ = \(\frac{2^3-1}{1-\frac{1}{2^3}}\) = \(\frac{2^3-1}{\frac{2^3-1}{2^3}}\) = 2³

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S_n = \(\frac{2^n-1}{1-\frac{1}{2^n}}\) = \(\frac{2^n-1}{\frac{2^n-1}{2^n}}\) = 2ⁿ

∴ S₁ + S₂ + S₃ + ..... + Sn = 2 + 2² + 2³ + ..... + 2ⁿ

= \(\frac{2(2^n-1)}{2-1}\) = 2(2ⁿ – 1)         \(\big(\because\,S_n=\frac{a(r^n-1)}{r-1},\text{Here}\,a=2.\,r=2\big).\)