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

Question

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

Infinity Learn · Accepted Answer

We find that  (1+x)2+(1+x)3+$$…$$(1+1)40+(1+mx)50=(1+x)2(1+x)48$$−$$1(1+x)−$$1+(1+mx)30=1x(1+x)50$$−(1+x)2+(1+mx)30$$∴ Coeffi. of $$x2$$ in $$(1+x)2+(1+x)3+$$…$$(1+x)49+(1+mx)50=$$ Coeffi. of $$x2$$ in $$1x(1+x)30$$−$$(1+x)2+(1+mx)50=$$ Coeffi. of $$x3$$ in $$(1+x)50$$−$$(1+x)2+$$ coefficient of $$x2$$ in $$(1+mx)50=50C3+51C2m2It$$ is given that this coefficient is $$(3n+1)51C3$$ ∴ $$50C3+50C2m2=(3n+1)51C3$$⇒ $$50$$×$$49$$×$$483$$×$$2$$×$$1+50$$×$$492$$×$$1m2=(3n+1)51$$×$$50$$×$$493$$×$$2$$×$$1$$⇒ $$16+m2=17(3n+1)⇒ $$m2=51n+1The$$ least value of n for which  $$51n+1$$ is a perfect square is $$5$$ and for this value of  n the value of  m is $$16$$. Hence, $$m=16$$ and  n $$=5$$.

Let $n$ be the smallest positive integer such that the coefficients $x^{2}$ in the expansion $(1 + x)^{2} + (1 + x)^{3} + \dots + (1 + x)^{49} + (1 + m x)^{50}$ is $(3 n + 1)^{51} C_{3}$ for some positive integer $n$ . Then the value of $n$ , is

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Let n be the smallest positive integer such that the coefficients x2 in the expansion (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

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Let $n$ be the smallest positive integer such that the coefficients $x^{2}$ in the expansion $(1 + x)^{2} + (1 + x)^{3} + \dots + (1 + x)^{49} + (1 + m x)^{50}$ is $(3 n + 1)^{51} C_{3}$ for some positive integer $n$ . Then the value of $n$ , is