The sum of n terms of the following series 1+(1+ x )+ [1+ x + x ^ 2 )+ . will be

Mathematics · JEE Main · NTA Exams — Progression and Series

1. \(\frac{1-\mathbf{x}^{n}}{1-\mathbf{x}}\)
2. \(\frac{\mathbf{x}\left(1-\mathbf{x}^{7}\right)}{1-\mathbf{x}}\)
3. \(\frac{n(1-x)-x\left(1-x^{n}\right)}{(1-x)^{2}}\)
4. None of the above

Answer

(C) n(1-x)-x (1-x^ n ) (1-x)^ 2

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