How do you find the coefficient of #x^2# in the expansion of #(2+x)^5#? Precalculus The Binomial Theorem Pascal's Triangle and Binomial Expansion 1 Answer Euan S. Aug 13, 2016 80 Explanation: Binomial theorem: #(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k)y^k# #(x+2)^5 = sum_(k=0)^5((5),(k)) x^(5-k)2^k# Looking for #x^2# so look at the #k=3# term: #((5),(3))x^2*2^3 = 8*(5!)/(3!2!)x^2 = 80x^2# Answer link Related questions What is Pascal's triangle? How do I find the #n#th row of Pascal's triangle? How does Pascal's triangle relate to binomial expansion? How do I find a coefficient using Pascal's triangle? How do I use Pascal's triangle to expand #(2x + y)^4#? How do I use Pascal's triangle to expand #(3a + b)^4#? How do I use Pascal's triangle to expand #(x + 2)^5#? How do I use Pascal's triangle to expand #(x - 1)^5#? How do I use Pascal's triangle to expand a binomial? How do I use Pascal's triangle to expand the binomial #(a-b)^6#? See all questions in Pascal's Triangle and Binomial Expansion Impact of this question 4949 views around the world You can reuse this answer Creative Commons License