If the constant term in the expansion of ( 3 x 2 - 1 2 x 5 ) 7 is, then is equal to _______ . [2023]

Question

If the constant term in the expansion of ${(3 x^{2} - \frac{1}{2 x^{5}})}^{7}$ is $α$ , then $[α]$ is equal to _______ . [2023]

Smart Achievers · Accepted Answer

(1275)

Let the $r^{t h}$ term be the constant term of the given expansion.

$∴$ $T_{r + 1} = {}^{7}{C_{r}} {(3 x^{2})}^{7 - r} \cdot {(\frac{- 1}{2 x^{5}})}^{r} = {}^{7}{C_{r}} 3^{7 - r} \cdot {(\frac{- 1}{2})}^{r} \cdot x^{14 - 2 r - 5 r}$

For constant term, put $14 - 2 r - 5 r = 0 \Rightarrow 7 r = 14 \Rightarrow r = 2$

$∴$ $T_{2 + 1} = {}^{7}{C_{2}} 3^{7 - 2} {(\frac{- 1}{2})}^{2} = 21 \times 3^{5} \times \frac{1}{4}$

$\Rightarrow α = 1275.75 \Rightarrow [α] = 1275$

Solution

Q.
If the constant term in the expansion of ${(3 x^{2} - \frac{1}{2 x^{5}})}^{7}$ is $α$ , then $[α]$ is equal to _______ . [2023]

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Solution

Q. If the constant term in the expansion of (3x2-12x5)7 is α, then [α] is equal to _______ . [2023]

Ans. (1275) Let the rth term be the constant term of the given expansion. ∴ Tr+1=Cr7(3x2)7-r·(-12x5)r=Cr737-r·(-12)r·x14-2r-5r For constant term, put 14-2r-5r=0⇒7r=14⇒r=2 ∴ T2+1=C2737-2(-12)2=21×35×14 ⇒α=1275.75⇒[α]=1275

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Q.
If the constant term in the expansion of ${(3 x^{2} - \frac{1}{2 x^{5}})}^{7}$ is $α$ , then $[α]$ is equal to _______ . [2023]