How do you expand this binomial using binomial theorem? #(2x+3)^10#

1 Answer
Oct 26, 2017

Answer:

#(2x+3)^10 = (2x)^10 + 10C1 (2x)^9*3 + 10C2 (2x)^8*3^2 + 10C3 (2x)^7 * 3^3 + 10C4 (2x)^6 * 3^4 + 10C5 (2x)^5 * 3^5 + 10C6 (2x)^6*3^6 + 10C7 (2x)^7*3^7 + 10C8 (2x)^8*3^8 + 10C9 (2x)^9*3^9 + 3^10#

Explanation:

#(2x+3)^10 = (2x)^10 + 10C1 (2x)^9*3 + 10C2 (2x)^8*3^2 + 10C3 (2x)^7 * 3^3 + 10C4 (2x)^6 * 3^4 + 10C5 (2x)^5 * 3^5 + 10C6 (2x)^6*3^6 + 10C7 (2x)^7*3^7 + 10C8 (2x)^8*3^8 + 10C9 (2x)^9*3^9 + 3^10#