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It Can Be Shown That (1+x)n=1+nx+n(n1)2!x2+n(n1)(n2)3!x3( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots

question 278

Multiple Choice

It can be shown that (1+x) n=1+nx+n(n1) 2!x2+n(n1) (n2) 3!x3( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots is true for any real number nn (not just positive integer values) and any real number xx , where x<1| x | < 1 . Use this series to approximate the given number to the nearest thousandth.
- 1.0631.06 ^ { 3 }


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