The term independent of x in the expansion of x2-1x12 is
Explanation:
We know:
(a + b)n = nC0 × a0 × bn + nC1 × a1 × bn-1 + nC2 × a2 × bn-2 + … + nCn × an × b0
Let the term independent of x in x2-1x12 is nCk × (x2)k × 1xn-k
= nCk × x2k × xk-n
= nCk × x3k–n
Now for this term to be independent of x, 3k – n = 0
⇒ k = n/3 = 12/3 = 4
∴ The term becomes 12C4.
Hence, option (c).
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