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Find the sum of the series1 + (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + ⋯ |
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Answer» Let Sn = 1 + (1 + x) + (1 + x + x2) + (1 + x + x2 + x3 ) + ⋯ to n terms Hence, (1 + x) Sn = (1 + x) + (1 − x)(1 + x) + (1 − x) + (1 + x + x2) + (1 − x) + (1 + x + x2 + x3) + ⋯ to n terms or (1 − x) Sn = (1 − x) + (1 − x2) + (1 − x3) to n terms + (1 − x4) + ⋯ = n − (x + x2 + x3 + x4 + ⋯ ) to n terms = n − \(\frac{x(1 − x^n)}{(1 − x)}\) Hence, Sn = \(\frac{n}{1-x}\) − \(\frac{x(1 − x^n)}{(1 − x)^2}\) |
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