binomial theorem

Question

Find the first negative term in the expansion of (1+x)^27/5 where x is a positive number.

Answer 1

Maybe you're talking about the binomial series?

 

(1+x)^k = 1 + kx + {k(k-1)}/2! x^2+.....+{k(k-1)(k-2)....(k-n+1)}/n! x^n

 

(1 + x)^27?5 = 1 + 27/5 x + (27/5*22/5)/2 x^2+.....

 

Your first negative term will be when you have 27/5-n+1 <0 or n> 32/5. The first term that fits this is the 7th term.

 

(27/5 * 22/5 * 17/5*12/5*7/5*2/5*(-3)/5 divided by 7!)( x^7)