Find the 7 th term in the expansion of 3 x2-1-x310-

T_n =T_r+1=( 1)^r x^n ry^r x ^10C_r n=7, r=6, x=3x^2, y =1/x^3 T_7 =T_6+1=( 1)^6 ^10C_6 (3x^2)^4 ( 1/x^3 )^6 =^10C_6 3^4 x^8 ×1/x^18=^10C_6×81/x^10=210 × 81/x^ ...

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Byju's Answer

Standard XII

Mathematics

General Term of Binomial Expansion

Find the 7 th...

Question

Find the 7th term in the expansion of ${(3 x^{2} - \frac{1}{x^{3}})}^{10} .$

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Solution

$T_{n} = T_{r + 1} = (- 1)^{r} x^{n - r} y^{r} x^{10} C_{r}$

$n = 7, r = 6, x = 3 x^{2}, y = \frac{1}{x^{3}}$

$T_{7} = T_{6 + 1} = (- 1)^{6}^{10} C_{6} (3 x^{2})^{4} {(\frac{1}{x^{3}})}^{6}$

$=^{10} C_{6} 3^{4} x^{8} \times \frac{1}{x^{18}} =^{10} C_{6} \times \frac{81}{x^{10}} = \frac{210 \times 81}{x^{10}}$

$= \frac{17010}{x^{10}}$

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Find the 7th term in the expansion of (3x2−1x3)10.

Tn=Tr+1=(−1)rxn−ryrx10Cr n=7,r=6,x=3x2,y=1x3 T7=T6+1=(−1)610C6(3x2)4(1x3)6 =10C634x8×1x18=10C6×81x10=210×81x10 =17010x10

Find the 7th term in the expansion of ${(3 x^{2} - \frac{1}{x^{3}})}^{10} .$