A. 3+9+5+... +(6n-3) = 3n²
B. 2+4+6+... +(2n) = n(n+1)
Jawab:
Penjelasan dengan langkah-langkah:
[tex]P_n :3+9+15 + ...+(6n-3) = 3n^2\\\text{Membuktikan ruas kanan dan kiri dengan mengambil sembarang nilai n}\\\text{n=1}\\6(1)-3 = 3(1)^2\\3 = 3 \implies \text{Sama}\\\\P_k \ \ \ : 3+9+15+...+(6k-3) = 3k^2\\P_{k+1} : 3+9+15+...+(6k-3)+(6k+3) = 3(k+1)^2\\\\3k^2+6k+3 = 3(k+1)^2\\3(k^2+2k+1) = 3(k+1)^2\\3(k+1)^2 = 3(k+1)^2 \implies \text{TERBUKTI}[/tex]
[tex]P_n : 2+4+6+...+2n = n(n+1)\\\text{Membuktikan ruas kanan dan kiri dengan mengambil sembarang nilai n}\\\text{n=1}\\2(1) = 1(1+1)\\2= 2 \implies \text{sama}\\\\P_k \ \ \ : 2+4+6+...+2k = k(k+1)\\P_{k+1} : 2 +4+6+...+2k+2(k+1) = (k+1)(k+2)\\\\k(k+1)+2k+2 = (k+1)(k+2)\\k^2+k+2k + 2= (k+1)(k+2)\\k^2+3k+2 = (k+1)(k+2)\\(k+1)(k+2) = (k+1)(k+2) \implies \text{TERBUKTI}[/tex]
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