determine the coefficient of the middle term in the expansion of (2x-3)^10

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember binomial expansion such that:

`(a+b)^n = C_n^0*a^n + C_n^1a^(n-1)*b^1 + ... + C_n^k*a^(n-k)*b^k + C_n^n*b^n`

Notice that there are n+1 terms in binomial expansion above, hence, if n=10, then there are 11 terms and there is one middle term `C_10^5*(2x)^5*3^5` .

Hence, the binomial coefficient is `C_10^5`  and the coefficient of variable x of the middle term is `C_10^5*(2*3)^5` .

`C_10^5 = (10!)/(5!*(10-5)!)`

`C_10^5 = (5!*6*7*8*9*10)/(5!*1*2*3*4*5)`

`C_10^5 = (7*2*9*2) = 252`

`C_10^5*(2*3)^5 = 252*7776 = 1959552`

Hence, evaluating the binomial coefficient of middle term of expansion `(2x-3)^10 ` yields `C_10^5 = 252`  and evaluating the  coefficient of variable x of  middle term of expansion `(2x-3)^10 ` yields `C_10^5*(2*3)^5 =1959552` .

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

We will use Pascal's Triangle to solve.

                                    1

                              1           1

                         1          2           1

                    1        3          3           1

                1     4         6           4            1

         1         5       10         10        5          1

...........................

1     10    45    120    210    252     210   120    45   10    1

 

Then the middle factor is 252.

`==gt (2X-3)^10 = (2X)^10+ 10(2X)^9 (-3) + ... + 252(2X)^5 (-3)^5`

`` ==> The middle term is `(252)(2x)^5 (-3)^5 = -1959552 x^5 `

``

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