How do you use the binomial series to expand # (1+x^6)^(1/2)#? Precalculus The Binomial Theorem The Binomial Theorem 1 Answer Shwetank Mauria Oct 17, 2016 #(1+x^6)^(1/2)=1+1/2x^6-1/8x^12+1/16x^18-5/128x^24+.......# Explanation: According to binomial series #(1+a)^n=1+na+(n(n-1))/(2!)a^2+(n(n-1)(n-2))/(3!)a^3+(n(n-1)(n-2)(n-3))/(4!)a^4+.......# Hence #(1+x^6)^(1/2)=1+(1/2)x^6+((1/2)(1/2-1))/(2!)x^12+((1/2)(1/2-1)(1/2-2))/(3!)x^18+((1/2)(1/2-1)(1/2-2)(1/2-3))/(4!)x^24+.......# or #(1+x^6)^(1/2)=1+1/2x^6+((1/2)(-1/2))/(2!)x^12+((1/2)(-1/2)(-3/2))/(3!)x^18+((1/2)(-1/2)(-3/2)(-5/2))/(4!)x^24+.......# or #(1+x^6)^(1/2)=1+1/2x^6-1/8x^12+1/16x^18-5/128x^24+.......# 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 2083 views around the world You can reuse this answer Creative Commons License