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Problem Which of the following power series equals sin x \sin x sin x x x x ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! \displaystyle \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!} n = 0 ∑ ∞ ( − 1 ) n ( 2 n )! x 2 n ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! \displaystyle \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!} n = 0 ∑ ∞ ( − 1 ) n ( 2 n + 1 )! x 2 n + 1 ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n ) ! \displaystyle \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n)!} n = 0 ∑ ∞ ( − 1 ) n ( 2 n )! x 2 n + 1 ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! \displaystyle \sum_{n=0}^{\infty}\dfrac{x^{2n+1}}{(2n+1)!} n = 0 ∑ ∞ ( 2 n + 1 )! x 2 n + 1 Browse the problem library