与えられた3つの式を展開せよ。 (1) $(x+y)^2$ (2) $(x+y)^3$ (3) $(x+y)^4$代数学展開二項定理多項式2025/5/251. 問題の内容与えられた3つの式を展開せよ。(1) (x+y)2(x+y)^2(x+y)2(2) (x+y)3(x+y)^3(x+y)3(3) (x+y)4(x+y)^4(x+y)42. 解き方の手順(1) (x+y)2(x+y)^2(x+y)2二項定理、または直接展開する。(x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2(x+y)2=x2+2xy+y2(2) (x+y)3(x+y)^3(x+y)3二項定理、または(x+y)2(x+y)^2(x+y)2の結果を使って展開する。(x+y)3=(x+y)(x+y)2=(x+y)(x2+2xy+y2)=x3+2x2y+xy2+x2y+2xy2+y3=x3+3x2y+3xy2+y3(x+y)^3 = (x+y)(x+y)^2 = (x+y)(x^2 + 2xy + y^2) = x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3 = x^3 + 3x^2y + 3xy^2 + y^3(x+y)3=(x+y)(x+y)2=(x+y)(x2+2xy+y2)=x3+2x2y+xy2+x2y+2xy2+y3=x3+3x2y+3xy2+y3(x+y)3=x3+3x2y+3xy2+y3(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3(x+y)3=x3+3x2y+3xy2+y3(3) (x+y)4(x+y)^4(x+y)4二項定理、または(x+y)3(x+y)^3(x+y)3の結果を使って展開する。(x+y)4=(x+y)(x+y)3=(x+y)(x3+3x2y+3xy2+y3)=x4+3x3y+3x2y2+xy3+x3y+3x2y2+3xy3+y4=x4+4x3y+6x2y2+4xy3+y4(x+y)^4 = (x+y)(x+y)^3 = (x+y)(x^3 + 3x^2y + 3xy^2 + y^3) = x^4 + 3x^3y + 3x^2y^2 + xy^3 + x^3y + 3x^2y^2 + 3xy^3 + y^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4(x+y)4=(x+y)(x+y)3=(x+y)(x3+3x2y+3xy2+y3)=x4+3x3y+3x2y2+xy3+x3y+3x2y2+3xy3+y4=x4+4x3y+6x2y2+4xy3+y4(x+y)4=x4+4x3y+6x2y2+4xy3+y4(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4(x+y)4=x4+4x3y+6x2y2+4xy3+y43. 最終的な答え(1) (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2(x+y)2=x2+2xy+y2(2) (x+y)3=x3+3x2y+3xy2+y3(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3(x+y)3=x3+3x2y+3xy2+y3(3) (x+y)4=x4+4x3y+6x2y2+4xy3+y4(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4(x+y)4=x4+4x3y+6x2y2+4xy3+y4