The problem asks us to calculate: ∑ i = 0 n ( − 1 ) i ( n i ) ( n n − i ) The way I tried solving is:The given sum is the coefficient of x n in ( 1 + x ) n ( 1 − x ) n , which is ( 1 − x 2 ) n The coefficient of xn in ( 1 − x 2 ) n is ( − 1 ) n / 2 ( n n / 2 ) .Am I doing it right?

Question

The problem asks us to calculate:      ∑          i      =      0          n    (  −  1      )    i                  (              n      i                  )                      (              n              n        −        i                        )      The way I tried solving is:The given sum is the coefficient of       x    n   in   (  1  +  x      )    n    (  1  −  x      )    n  , which is   (  1  −      x    2        )    n   The coefficient of xn in   (  1  −      x    2        )    n   is  (  −  1      )          n          /      2                      (              n              n              /        2                        )        .Am I doing it right?

Marlee Guerra · Accepted Answer

Your solution is correct for even   n.If   n is odd then your last sentence should read &quot;The coefficient of       x    n   in   (  1  −      x    2        )    n   is   0. This is because only even powers of   x occur when expanding   (  1  −      x    2        )    n  .

gvaldytist · Accepted Answer

∑                          k              =              0                                  n                                              (                            −              1                            )                                  k                                                                  (                                      n              k                                                  )                                                                              (                                      n                              n                −                k                                                                )                                      =                  ∑                      k            =            0                            n                                      (                        −            1                        )                            k                                                      (                                n            k                                          )                                                                                               ∮                                                      |                                        z                                        |                                    =                                      1                                      −                                                                                                                                (                                        1                    +                    z                                        )                                                    n                                                  z                                      n                    −                    k                    +                    1                                                                                                                                          d                  z                                                  2                  π                                  i                                                      ⏞                                                      =                                                                                           (                                                  n                                      n                    −                    k                                                                                    )                                                                                        =                                       ∮                                    |                            z                            |                        =                          1                          −                                                                                (                            1              +              z                            )                                  n                                z                          n              +              1                                                                                                           ∑                                  k                  =                  0                                              n                                                                                          (                                                  n                  k                                                                  )                                                                              (                                    −                  z                                    )                                              k                                      ⏞                                                      =                                                               (                                    1                  −                  z                                    )                                              n                                                                                                           d            z                                2            π                      i                          =                  ∮                                    |                            z                            |                        =                          1                          −                                                                                (                            1              −                              z                              2                                          )                                  n                                z                          n              +              1                                                                                    d            z                                2            π                      i                                              =                                       ∑                      k            =            0                            n                                                      (                                n            k                                          )                                                (                        −            1                        )                            k                                                                                     ∮                                                      |                                        z                                        |                                    =                                      1                                      −                                                                                          1                                  z                                      n                    −                    2                    k                    +                    1                                                                        ⏟                                                      =                                             δ                                  n                  ,                  2                  k                                                                                                                  d            z                                2            π                      i                          =                              {                                                                                                                                                        (                                                    −                          1                                                    )                                                                          n                                                  /                          2                                                                                                                                                                  (                                                                          n                                                      n                                                      /                            2                                                                                                                            )                                                                                                                                                          if                                                                                                          n                                                               is                                                             e                    v                    e                    n                                                                                                                                      0                                                                                                        otherwise

The problem asks us to calculate: <munderover> ∑ i =

Answered question

Answer & Explanation

New Questions in Precalculus