Question
If the coefficient of x in the expansion of (x2+kx)5 is 270, then k =
Answer: Option C
:
C
Tr+1 =5Cr(x2)5−r(kx)r
For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3
Hence, T3+1 =5C3(x2)5−3(kx)3
According to question, 10k3 = 270 ⇒ k = 3.
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:
C
Tr+1 =5Cr(x2)5−r(kx)r
For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3
Hence, T3+1 =5C3(x2)5−3(kx)3
According to question, 10k3 = 270 ⇒ k = 3.
Was this answer helpful ?
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