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The coefficient of $x^4$ in the expansion of $(1 + 2x)^6$ is
A$\binom{6}{4} = 15$
B$4!\,\binom{6}{4}$
C$2^4\,\binom{6}{4} = 240$
D$2^4\,\binom{6}{2} = 240$
Answer & Solution
Correct answer: C. $2^4\,\binom{6}{4} = 240$
1. General term of $(1 + 2x)^6$: $T_{r+1} = \binom{6}{r}\,(2x)^r = \binom{6}{r}\,2^r\,x^r$.
2. For $x^4$ coefficient, set $r = 4$.
3. Coefficient = $\binom{6}{4} \cdot 2^4 = 15 \cdot 16 = 240$.
4. Option A forgets the $2^r$ factor (would be the coefficient of $x^4$ in $(1+x)^6$, not $(1+2x)^6$). Option C uses $\binom{6}{2} = 15$, but that's the symmetric value of $\binom{6}{4}$ — same number 15, so numerically equivalent to A's 15 (but stated as different binomial). Option D has wrong factor.
5. Note B and C have the same final numerical value 240 but different notation; B is the canonical form.
_Source: NCERT Class 11 Mathematics, Ch 7, §7.2.1 (Examples), p. 5–7._
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