The coefficient of x35 in the expansion of (1+x5)10 is

Question

The coefficient of x35 in the expansion of (1+x5)10 is (A) 120 (B) 210 (C) 240 (D) 480

Tardigrade Admin · Accepted Answer

T r +1= 10 C r ⋅ x 5 r ; r =0,1,2, ldots ldots ., 10 ∴ For coefficient of x35, put 5 r=35 ∴ r =7 So, T 8= 10 C 7 x 35 ⇒ Coefficient of x 35= 10 C 7=(10 !/7 ! ⋅ 3 !)=(10 × 9 × 8/6)=120

Q. The coefficient of $x^{35}$ in the expansion of $(1 + x^{5})^{10}$ is

120

210

240

480

$T_{r + 1} =^{10} C_{r} \cdot x^{5 r}; r = 0, 1, 2, \dots\dots ., 10$
$∴$ For coefficient of $x^{35}$ , put $5 r = 35$
$∴ r = 7$
So, $T_{8} =^{10} C_{7} x^{35} \Rightarrow$ Coefficient of $x^{35} =^{10} C_{7} = \frac{10 !}{7 ! \cdot 3 !} = \frac{10 \times 9 \times 8}{6} = 120$

Q. The coefficient of x35 in the expansion of (1+x5)10 is

120

210

240

480

Tr+1​=10Cr​⋅x5r;r=0,1,2,…….,10 ∴ For coefficient of x35, put 5r=35 ∴r=7 So, T8​=10C7​x35⇒ Coefficient of x35=10C7​=7!⋅3!10!​=610×9×8​=120

Q. The coefficient of $x^{35}$ in the expansion of $(1 + x^{5})^{10}$ is

$T_{r + 1} =^{10} C_{r} \cdot x^{5 r}; r = 0, 1, 2, \dots\dots ., 10$
$∴$ For coefficient of $x^{35}$ , put $5 r = 35$
$∴ r = 7$
So, $T_{8} =^{10} C_{7} x^{35} \Rightarrow$ Coefficient of $x^{35} =^{10} C_{7} = \frac{10 !}{7 ! \cdot 3 !} = \frac{10 \times 9 \times 8}{6} = 120$