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Binomial expansion Find Constant term / independent term using Binomial expansion method Examples ( Enter your problem )
  1. Expand all terms using Binomial theorem method Examples
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  4. Expand `x^n` term using Binomial expansion method Examples
  5. Find Middle term using Binomial expansion method Examples
  6. Find Constant term / independent term using Binomial expansion method Examples
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5. Find Middle term using Binomial expansion method Examples
(Previous example)

6. Find Constant term / independent term using Binomial expansion method Examples





1. Find Constant term / independent term of `(2-3x)^5`

Solution:
`(2-3x)^5`

Suppose the constant term exists and it is `(r+1)^(th)` term

`"Here "a=2,b=-3x,n=5`

`T_(r+1)=((n),(r))a^(n-r)b^r`

`=((5),(r))(2)^(5-r)(-3x)^r`

`=((5),(r))(2)^(5-r)*(-3)^r*(x)^r`

`=((5),(r))(2)^(5-r)*(-3)^r*(x)^(r)`

Determine the value of `r` for the constant term

For the constant term, the power of `x` must be 0

`r=0`

To find `T_1`, we let `r=0`

`:. T_1=T_(0+1)`

`=((5),(0))(2)^5(-3x)^0`

`=1(2)^5(-3x)^0`

`=1(32)(1)`

`=32`

The constant term in the expansion of `(2-3x)^5` is `32`
2. Find Constant term / independent term of `(2x^2-3)^11`

Solution:
`(2x^2-3)^11`

Suppose the constant term exists and it is `(r+1)^(th)` term

`"Here "a=2x^2,b=-3,n=11`

`T_(r+1)=((n),(r))a^(n-r)b^r`

`=((11),(r))(2x^2)^(11-r)(-3)^r`

`=((11),(r))(2)^(11-r)*(x^2)^(11-r)*(-3)^r`

`=((11),(r))(2)^(11-r)*(-3)^r*(x^2)^(11-r)`

`=((11),(r))(2)^(11-r)*(-3)^r*(x)^(2(11-r))`

`=((11),(r))(2)^(11-r)*(-3)^r*(x)^(22-2r)`

Determine the value of `r` for the constant term

For the constant term, the power of `x` must be 0

`22-2r=0`

`2r=22`

`r=11`

To find `T_12`, we let `r=11`

`:. T_12=T_(11+1)`

`=((11),(11))(2x^2)^0(-3)^11`

`=1(2x^2)^0(-3)^11`

`=1(1)(-177147)`

`=-177147`

The constant term in the expansion of `(2x^2-3)^11` is `-177147`
3. Find Constant term / independent term of `(2x^2-1/x)^11`

Solution:
`(2x^2-1/x)^11`

Suppose the constant term exists and it is `(r+1)^(th)` term

`"Here "a=2x^2,b=-1/x,n=11`

`T_(r+1)=((n),(r))a^(n-r)b^r`

`=((11),(r))(2x^2)^(11-r)(-1/x)^r`

`=((11),(r))(2)^(11-r)*(x^2)^(11-r)*(-1)^r*(1/x)^r`

`=((11),(r))(2)^(11-r)*(-1)^r*(x^2)^(11-r)*(1/x)^r`

`=((11),(r))(2)^(11-r)*(-1)^r*(x)^(2(11-r))*(x)^(-r)`

`=((11),(r))(2)^(11-r)*(-1)^r*(x)^(22-2r-r)`

`=((11),(r))(2)^(11-r)*(-1)^r*(x)^(22-3r)`

Determine the value of `r` for the constant term

For the constant term, the power of `x` must be 0

`22-3r=0`

`3r=22`

`r=22/3`

Since `r` must be an integer, there is no term where the power of `x` is 0

constant term does not exist in the expansion
4. Find Constant term / independent term of `(2x^2-1/x^2)^11`

Solution:
`(2x^2-1/x^2)^11`

Suppose the constant term exists and it is `(r+1)^(th)` term

`"Here "a=2x^2,b=-1/(x^2),n=11`

`T_(r+1)=((n),(r))a^(n-r)b^r`

`=((11),(r))(2x^2)^(11-r)(-1/(x^2))^r`

`=((11),(r))(2)^(11-r)*(x^2)^(11-r)*(-1)^r*(1/(x^2))^r`

`=((11),(r))(2)^(11-r)*(-1)^r*(x^2)^(11-r)*(1/(x^2))^r`

`=((11),(r))(2)^(11-r)*(-1)^r*(x)^(2(11-r))*(x)^(-2r)`

`=((11),(r))(2)^(11-r)*(-1)^r*(x)^(22-2r-2r)`

`=((11),(r))(2)^(11-r)*(-1)^r*(x)^(22-4r)`

Determine the value of `r` for the constant term

For the constant term, the power of `x` must be 0

`22-4r=0`

`4r=22`

`r=11/2`

Since `r` must be an integer, there is no term where the power of `x` is 0

constant term does not exist in the expansion




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5. Find Middle term using Binomial expansion method Examples
(Previous example)





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