1.The coefficient of xnx^nxn in the expansion of loga(1+x)\log_a (1 + x)loga(1+x) isa.(−1)n−1n\frac{(-1)^{n-1}}{n}n(−1)n−1b.(−1)n−1nlogae\frac{(-1)^{n-1}}{n \log_a e}nlogae(−1)n−1c.(−1)n−1nlogea\frac{(-1)^{n-1}}{n \log_e a}nlogea(−1)n−1d.(−1)nnlogae\frac{(-1)^n}{n \log_a e}nlogae(−1)nLogin to continueOnly logged in users canattempt or see the solution.
