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

*print*Print*list*Cite

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` .**

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 `**

**`` **