If the constant term in the binomial expansion of ( x-kx^2)^10 is 405, then |k| equals:

Question

If the constant term in the binomial expansion of $\left(\sqrt x-\dfrac{k}{x^2}\right)^{10}$ is 405, then $|k|$ equals:

Accepted Answer

$${\left( {\sqrt x  - {k \over {{x^2}}}} \right)^{10}}$$

rth term of the expansion,

Tr+1 = 10Cr$${\left( {\sqrt x } \right)^{10 - r}}{\left( {{{ - k} \over {{x^2}}}} \right)^r}$$

= 10Cr.$${x^{{{10 - r} \over 2}}}.{\left( { - k} \right)^r}.{x^{ - 2r}}$$

= 10Cr.$${x^{{{10 - 5r} \over 2}}}.{\left( { - k} \right)^r}$$

If it is constant term then
$${{{10 - 5r} \over 2}}$$ = 0

$$ \Rightarrow $$ r = 2

T3 = 10C2.(-k)2 = 405

$$ \Rightarrow $$ k2 = $${{405} \over {45}}$$ = 9

$$ \Rightarrow $$ k = $$ \pm $$ 3

$$ \Rightarrow $$ |k| = 3

Solutions · 1