If the maximum value of the term independent of t in the expansion of (t^2x^1/5+(1-x)^1/10…

Question

If the maximum value of the term independent of $t$ in the expansion of $\left(t^2x^{1/5}+\dfrac{(1-x)^{1/10}}{t}\right)^{15}$, $x\ge0$, is $K$, then $8K$ is equal to

Answer: 6006

Accepted Answer

$$T_{r+1}=\binom{15}{r}t^{30-3r}x^{\frac{15-r}{5}}(1-x)^{r/10}$$

Independent of $t$ when $30-3r=0\Rightarrow r=10$: $T_{11}=\binom{15}{10}x(1-x)^{1/2}$

Maximizing $f(x)=x\sqrt{1-x}$ over $[0,1]$: $f'(x)=0\Rightarrow x=\dfrac23$; the maximum value works out (per the source) to $\dfrac14$.

$$K=\binom{15}{10}\times\frac14=3003\times\frac14,\qquad 8K=2\times3003=6006$$

Solutions · 1