How do you use the binomial series to expand #(1+x^2)^(2/3)#? Precalculus The Binomial Theorem The Binomial Theorem 1 Answer Cesareo R. Nov 19, 2016 #(1+x^2)^(2/3)=sum_(k=0)^oo(Pi_(j = 1)^k(2/3 - j + 1))/(k!)x^(2k)# convergent for #absx lt 1# Explanation: #(1+a)^n=1+na+(n(n-1))/(2!)a^2+cdots+(Pi_(j=k)^n (n-j+1))/(k!)a^k# so #(1+x^2)^(2/3)=sum_(k=0)^oo(Pi_(j = 1)^k(2/3 - j + 1))/(k!)x^(2k)# convergent for #absx lt 1# Answer link Related questions What is the binomial theorem? How do I use the binomial theorem to expand #(d-4b)^3#? How do I use the the binomial theorem to expand #(t + w)^4#? How do I use the the binomial theorem to expand #(v - u)^6#? How do I use the binomial theorem to find the constant term? How do you find the coefficient of x^5 in the expansion of (2x+3)(x+1)^8? How do you find the coefficient of x^6 in the expansion of #(2x+3)^10#? How do you use the binomial series to expand #f(x)=1/(sqrt(1+x^2))#? How do you use the binomial series to expand #1 / (1+x)^4#? How do you use the binomial series to expand #f(x)=(1+x)^(1/3 )#? See all questions in The Binomial Theorem Impact of this question 1276 views around the world You can reuse this answer Creative Commons License