First slide

Binomial theorem for positive integral Index

Question

If $f (x) =^{40} C_{1} \cdot x (1 - x)^{39} + 2 \cdot^{40} C_{2} x^{2} (1 - x)^{38} + 3 \cdot^{40} C_{3} x^{3} (1 - x)^{37} + \dots + 40 \cdot^{40} C_{40} x^{40}$ , then the value of f(3) is

Moderate

A

120
B

150
C

200
D

240

Solution

$f (x) =^{40} C_{1} \cdot x (1 - x)^{39} + 2 \cdot^{40} C_{2} \times x^{2} (1 - x)^{38} + 3 \cdot^{40} C_{3} \times x^{3} (1 - x)^{37} + \dots + 40 \cdot^{40} C_{40} \times x^{40}$
$\begin{matrix} T_{r} = r \cdot^{40} C_{r} \cdot x^{r} (1 - x)^{40 - r} \\ = 40 x \cdot^{39} C_{r - 1} \cdot x^{r - 1} (1 - x)^{40 - r} \\ ∴ f (x) = 40 x (x + 1 - x)^{39} \\ = 40 x \end{matrix}$

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