Jika S₁, S₂, dan S₃ merupakan jumlah n suku, 2n suku, dan 3n suku pertama dari deret geometri, buktikan bahwa: a. S₁(S₃ - S₂) = (S₂ - S₁)² b. S₁² + S₂² = S₁(S₂

Materi: Barisan dan Deret.

[tex]\displaystyle S_1(S_3-S_2)=S_1\left(\frac{a(r^{3n}-1)}{r-1}-\frac{a(r^{2n}-1)}{r-1}\right)\\S_1(S_3-S_2)=S_1\left(\frac{a(r^{n}-1)(1+r^n+r^{2n})}{r-1}-\frac{a(r^{n}-1)(r^n+1)}{r-1}\right)\\S_1(S_3-S_2)=S_1\left(\frac{a(r^{n}-1)}{r-1}(1+r^n+r^{2n}-r^n-1)\right)\\S_1(S_3-S_2)=S_1\left(\frac{a(r^{n}-1)}{r-1}(r^{2n})\right)\\S_1(S_3-S_2)=\frac{a(r^{n}-1)}{r-1}\left(\frac{a(r^{3n}-r^{2n})}{r-1}\right)\\S_1(S_3-S_2)=\frac{a^2(r^{4n}-r^{3n}-r^{3n}+r^{2n})}{(r-1)^2}[/tex]
[tex]\displaystyle S_1(S_3-S_2)=\frac{a^2(r^{4n}-2r^{3n}+r^{2n})}{(r-1)^2}\\S_1(S_3-S_2)=\frac{a^2(r^{4n}-2r^{2n}\cdot r^n+r^{2n})}{(r-1)^2}\\S_1(S_3-S_2)=\frac{a^2(r^{2n}-r^n)^2}{(r-1)^2}\\S_1(S_3-S_2)=\frac{a^2(r^{2n}-1-(r^n+1))^2}{(r-1)^2}\\S_1(S_3-S_2)=\left(\frac{a(r^{2n}-1)-a(r^n+1)}{r-1}\right)^2\\S_1(S_3-S_2)=\left(\frac{a(r^{2n}-1)}{r-1}-\frac{a(r^{n}-1)}{r-1}\right)^2\\\boxed{\boxed{S_1(S_3-S_2)=\left(S_2-S_1\right)^2}}[/tex]


[tex]\displaystyle S_1(S_2+S_3)=S_1\left(\frac{a(r^{2n}-1)}{r-1}+\frac{a(r^{3n}-1)}{r-1}\right)\\S_1(S_2+S_3)=\frac{a(r^{n}-1)}{r-1}\left(\frac{a(r^{3n}+r^{2n}-2)}{r-1}\right)\\S_1(S_2+S_3)=\frac{a^2(r^{4n}+r^{3n}-2r^n-r^{3n}-r^{2n}+2)}{(r-1)^2}\\S_1(S_2+S_3)=\frac{a^2(r^{4n}-r^{2n}-2r^n+2)}{(r-1)^2}\\S_1(S_2+S_3)=\frac{a^2(r^{4n}-2r^{2n}+1+r^{2n}-2r^n+1)}{(r-1)^2}\\S_1(S_2+S_3)=\frac{a^2((r^{2n}-1)^2+(r^{n}-1)^2)}{(r-1)^2}\\S_1(S_2+S_3)=\frac{a^2(r^{2n}-1)^2+a^2(r^{n}-1)^2}{(r-1)^2}[/tex]
[tex]\displaystyle S_1(S_2+S_3)=\frac{a^2(r^{2n}-1)^2}{(r-1)^2}+\frac{a^2(r^{n}-1)^2}{(r-1)^2}\\\boxed{\boxed{S_1(S_2+S_3)=S_2^2+S_1^2}}[/tex]