The sum of the binomial coefficients in the expansion of (x2+1/x)n is 1024 Find the coefficient of x11 in - Maths - Binomial Theorem

Question

The sum of the binomial coefficients in the expansion of
(x2+1/x)n is 1024 Find the coefficient of
x11 in the binomial expansion

Ishwarmani · Accepted Answer

Dear Student, Please find below the solution to the asked query:   Expression : x2+1xnsum of coefficients = 1024⇒2n= 1024⇒n= 10Suppose r+1th term contains x11 in the expansion of x2+1x10tr+1 = Cr10×x2r ×1x10-r = Cr10×x3r-10   
1If tr+1 contains x11, then3r-10=11⇒r = 7Now, from 1, we gett8 = C710×x11 = 120x11So, coefficient of x11 = 120Hope this information will clear your doubts about the topic
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The sum of the binomial coefficients in the expansion of (x2+1/x)n is 1024. Find the coefficient of x11 in the binomial expansion.

The sum of the binomial coefficients in the expansion of (x²+1/x)ⁿ is 1024. Find the coefficient of x¹¹ in the binomial expansion.