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ruas kiri : 4n = 4(1) = 4
ruas kanan = 2n(n+1) ÷ 2(1)(1+1) = 4
benar
2) andaikan benar untuk n = k,
4+8+12+16+...+ 4k = 2k(k+1)
3) maka harus memenuhi untuk n = k + 1,
4+8+12+16+...+4k + 4(k+1) = 2(k+1)(k+1+1)
2k(k+1) + 4(k+1) = 2(k+1)(k+2)
2k^2 + 2k + 4k + 4 = 2(k^2 + 3k + 2)
2k^2 + 6k + 4 = 2k^2 + 6k + 4
terbukti!