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Mathematics Term 1  STPM  Chapter 2 Sequences and Series
                                            n
               The expansion of (1 + x) , n  Q
                                                                       +
                                                           n
               We have seen that the binomial expansion of (1 + x) , where n ∈ Z , is a finite series with (n + 1) terms, i.e.
                                                                                      n
                                 (1 + x)  = 1 + nx +   n(n – 1)   x  +  n (n – 1)(n – 2)  x  +  …  + x .
                                                           2
                                                                             3
                                       n
                                                     2!              3!
               However, if n is any rational number, i.e. n ∈ Q, then the expansion is an infinite series, i.e.
                     (1 + x) = 1 + nx +   n(n – 1)   x  +   n(n – 1)(n – 2)  x  +  …  +   n(n – 1) … (n – r + 1)   x  +  …
                                                                 3
                                               2
                           n
                                                                                              r
                                         2!             3!                        r!
               This series is called the binomial series, and is valid if |x| , 1, i.e. –1 , x , 1. It is used to find approximations,
               up to a degree of accuracy very close to its actual value.                                   2
                                 n
                                          n!
               Note:  The notation  1 2  =   (n – r)! r!  is not applicable if n is not a positive integer.
                                  r
                   Example 38

                 Expand each of the following expressions as an ascending series in x, up to the term in x . State the range
                                                                                          4
                 of x such that the expansion is valid.
                (a)   1 1 +   1  x 2 — 1 3                           (b)  (1 – 2x) –3
                         2

                Solution:           By using the binomial expansion
                                                              2
                                                                                3
                                    (1 + x)  = 1 + nx +   n(n – 1)  x  +  n(n – 1)(n – 2)  x
                                          n
                                                         2!            3!
                                              +   n(n – 1)\(n – 2)(n – 3)   x  +  …
                                                                     4
                                                         4!
                                    (a)  Substitute x =   1   x and n =   1  , we get
                                                     2          3
                                                                 1 1  – 1 2	       1 1  – 1 21  1  – 2 2
                                                                                    1
                                                                  1


                                                 1
                                                —
                                                2
                                                                              2
                                                                                                      2
                                                 3
                                                          1 2
                                                                                                       3
                                                                              2
                                         1 1 +   1  x  = 1 +   1 1 x  +   3 3 2!    1 1  x  +   3 3  3!  3    1 1  x

                                                                           2
                                                                                                   2
                                                        3 2
                                             2
                                                       1 1  – 1 21  1  – 2 21  1  – 3 2
                                                         1

                                                                                 2
                                                    +   3 3     3      3       1 1  x  +  …
                                                                                 4
                                                                 4!           2
                                                              1   –  2 2     1   –  2 21 –  5 2
                                                                               1
                                                                1
                                                                          2
                                                                                             2
                                                  = 1 +   1  x +   3  3    1 1   x  +   3  3  3    1 1  x
                                                                                             3
                                                                         2
                                                        6       2     4           6       8
                                                          1    –  2 21 –  5 21 –  8 2
                                                           1
                                                          3   3   3   3    1
                                                                              2
                                                                              4
                                                    +                      1   x  +  …
                                                                24        16
                                                  = 1 +   1 x –   1  x  +   5  x  –   5  x  +  …
                                                                2
                                                                       3
                                                                                4
                                                        6    36    648     1944
                                                               1
                                         The expansion is valid if |   x| , 1, i.e. |x| , 2 or –2 , x , 2.

                                                               2
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       02 STPM Math T T1.indd   127                                                                    3/28/18   4:21 PM
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