Let m be the smallest positive integer such that the coefficient of x² in the expansion of (1+x)²+(1+x)³+...+(1+x)⁴⁹+(1+mx)⁵⁰ is (3n+1)⁵¹C₃ for some positive integer n. Then the value of n is.<p></p>

Coeff. ofx2:2C2+3C2+4C2+..+49C2+50C2m2=(3n+1)51C3⇒3C3+3C2+4C2+..+49C2+50C2m2=(3n+1)51C3⇒50C3+50C2.m2=(3n+1)51C3(∵nCr+nCr𕒵=n+1Cr)⇒50C3+50C2+(m2𕒵)50C2=3n.513.50C2+51C3⇒51C3+(m2𕒵)50C2=51n.50C2+51C3⇒m2𕒵=51n⇒m2=51n+1Forn=5,mwill be the smallest positive integer.

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Question:

Let m be the smallest positive integer such that the coefficient of x² in the expansion of (1+x)²+(1+x)³+...+(1+x)⁴⁹+(1+mx)⁵⁰ is (3n+1)⁵¹C₃ for some positive integer n. Then the value of n is.

Solution: