IndukSI Matematika
i. n =1
ii. n = k
iii, n = k + 1
Penjelasan dengan langkah-langkah:
induksi dari
2 + 7 + 12 + . . . + (5n - 3) = [tex]\sf \frac{1}{2}n(5n-1)[/tex]
i. n = 1
5(1) - 3 = [tex]\sf \frac{1}{2}(1)(5-1)[/tex]
2= 2
ii. n = k
[tex]\sf 2+ 7 + 12 + . . . +(5k-3) = \frac{1}{2} k(5k -1)[/tex]
iii. n = k + 1
[tex]\sf 2+ 7 + 12 + . . . +(5k-3) + \{5(k+1) - 3\} = \frac{1}{2} (k+1)\{5(k+1) -1\}[/tex]
[tex]\sf \frac{1}{2}k(5k-1) + \{5(k+1) - 3\} = \frac{1}{2} (k+1)\{5(k+1) -1\}[/tex]
[tex]\sf \frac{1}{2}k(5k-1) +(5k+5 - 3) = \frac{1}{2} (k+1)(5k+5 -1)[/tex]
[tex]\sf \frac{1}{2}(5k^2-k) +(5k+2) = \frac{1}{2} (k+1)(5k+4)[/tex]
[tex]\sf \frac{1}{2}\{(5k^2-k) +2(5k+2)\} = \frac{1}{2} (k+1)(5k+4)[/tex]
[tex]\sf \frac{1}{2}(5k^2-k+10k+4) = \frac{1}{2} (k+1)(5k+4)[/tex]
[tex]\sf \frac{1}{2}(5k^2+9k+4) = \frac{1}{2} (k+1)(5k+4)[/tex]
[tex]\sf \frac{1}{2}(k+1)(5k+4)= \frac{1}{2} (k+1)(5k+4)[/tex]
ruas kiri = ruas kanan
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