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📐 Mathematics  ·  Binomial Theorem  ·  JEE

Find the term containing x³ in (3 + x/2)⁸.

Answer: 8C3 × 3⁵ × x³/8.

  • A 56 × 3⁵/8
  • B 7C3 x 3<sup>5</sup> x (x/2)<sup>3</sup>
  • C 8C3 × 3⁵ × x³/8
  • D 8C5 × 3⁵ × x³

Correct answer: C. 8C3 × 3⁵ × x³/8

Explanation: T(r+1) = 8Cr x 3<sup>8-r</sup> x (x/2)<sup>r.</sup> For x<sup>3</sup>: r=3. T<sub>4</sub> = 8C3 x 3<sup>5</sup> x x<sup>3</sup>/8 = 56 x 243 x x<sup>3</sup>/8.

Pascal's Triangle: Binomial Coefficients11112113311464115101051n=0n=1n=2n=3n=4n=5Each entry = sum of the two entries diagonally above it (Pascal's identity); row n gives nC0...nCn

Row n of Pascal's triangle gives the coefficients nC0, nC1, ..., nCn for the expansion of (a+b)ⁿ; each number is the sum of the two numbers diagonally above it, a direct visual proof of the identity nCr = (n-1)C(r-1) + (n-1)Cr.

Concept context

Expansion of (a+b) n , general term, middle term, binomial coefficients, and greatest term

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