Let f(x)= 1-x^ n+1 1-x and g (x) = 1 – 2 x + 3 x^ 2 - .........+ (–1) n n +1 x ^ n . Then…
Mathematics · JEE Advanced · NTA Exams — Limit, Continuity and Differentiability
Let \(f(x)=\frac{1-x^{n+1}}{1-x}\)and g (x) = 1 –\(\frac{2}{x}+\frac{3}{x^{2}}-\).........+ (–1)n \(\frac{\mathrm{n}+1}{\mathrm{x}^{\mathrm{n}}}\). Then the constant term in f’(x) × g(x) is equal to
- \(\frac{\mathrm{n}\left(\mathrm{n}^{2}-1\right)}{6}\) when n is even
- \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\) when n is odd
- \(-\frac{\mathrm{n}}{2}(\mathrm{n}+1)\) when n is even
- \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) when n is odd
Answer
(B) n ( n +1) 2 when n is odd, (C) - n 2 ( n +1) when n is even
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