JEE MAIN - Mathematics (2003 - No. 40)
If $$x$$ is positive, the first negative term in the expansion of $${\left( {1 + x} \right)^{27/5}}$$ is
6th term
7th term
5th term
8th term.
Giải thích
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.
$$\therefore$$ 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.