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MISCELLANEOUS EXERCISE 4 (II) · Q128

Q.The 3rd term of (1+x)n(1+x)^n is 36x236x^2. Find 5th term.

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3rd term of (1+x)n(1+x)^n is t3=nC2x2t_3={}^nC_2x^2. Given nC2=36^nC_2=36: n(n−1)2=36⇒n2−n−72=0⇒(n−9)(n+8)=0⇒n=9\dfrac{n(n-1)}2=36\Rightarrow n^2-n-72=0\Rightarrow(n-9)(n+8)=0\Rightarrow n=9 (rejecting n=−8n=-8). 5 …

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