Posted by **anoynomous** on Friday, March 22, 2013 at 2:22pm.

how can we show that there is no constant

in the expansion of [2x – (x^2/4)]^9

- maths-binomial theorem -
**Reiny**, Friday, March 22, 2013 at 2:34pm
general term

= C(9,n) (2x)^(9-n) (x^2/4)^n

= C(9,n) 2^(9-n) x^(9-n) (1/4)^n x^2n)

= C(9,n) (2^(9-n)(1/4)^n x^(9+n)

to have a constant, the exponent of x^(9+n) must be zero

so 9+n = 0

n = -9

BUT, n must be a positive integer, so there is no constant term in the expansion.

## Answer This Question

## Related Questions

- maths - in the binomial expansion of (1+x/k)^n, where k is a constant and n is a...
- maths-binomial theorem - coefficient of x^39 in the expansion of [(1/x^2) + x^4...
- maths-binomial theorem--plse help - find the greatest term in the expansion of (...
- maths - Use binomial theorem to to expand squareroot of 4+x in ascending powers ...
- maths - binomial theorem - Write and proof binomial theorem for rational indices...
- Algebra Binomial Theorem - Hi, can somebody help me with this math problem I am ...
- maths - find the first 3 terms in the expansion of (x+4)(1+3x)^-2 as a series in...
- binomial theorem - find the 20th term in the expansion of (a+b)^22
- Integrated Math 1 - What is the coefficient of the xy^2 term in the expansion of...
- binomial expansion - how do you expand the following as a series of ascending ...

More Related Questions