If x is positive, the first negative term in the expansion of (1+x)^27/5 is:

Question

If $x$ is positive, the first negative term in the expansion of $(1+x)^{27/5}$ is:

Accepted Answer

General term of $${\left( {1 + x} \right)^{n}}$$ is (${T_{r + 1}}$) = $${{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}$$

$$\therefore$$ General term of $${\left( {1 + x} \right)^{27/5}}$$ = $${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r + 1} \right)} \over {1.2.3....r}}{x^r}$$

For first negative term, $${\left( {{{27} \over 5} - r + 1} \right)}$$ < 0

$$ \Rightarrow r > {{27} \over 5} + 1$$

$$ \Rightarrow r > {{32} \over 5}$$

$$ \Rightarrow r > 6.4$$

$$\therefore$$ r = 7

${T_{7 + 1}} = {T_8}$ means 8th term is the first negative term.

Solutions · 1