Penjelasan dengan langkah-langkah:

Nomor 1

2+4+6+...+2n = n(n+1)

• Untuk n = 1

2n = n(n+1)

2.1 = 1.(1+1)

2 = 1.2

2 = 2 Benar

• Asumsikan n = k, maka n = k+1 bernilai benar juga.

2+4+6+...+2n = n(n+1)

2+4+6+...+2k = k(k+1)

(2+4+6+...+2k) + 2.(k+1) = (k+1)(k+1+1)

k(k+1) + 2k + 2 = (k+1)(k+2)

k² + k + 2k + 2 = k² + 2k + k + 2

k² + 3k + 2 = k² + 3k + 2 Benar

Nomor 2

1+4+7+...+(3n-2) = n(3n-1)/2

• Untuk n = 1

(3.1 - 2) = 1.(3.1-1)/2

3 - 2 = 1.(3-1)/2

1 = 2/2

1 = 1 Benar

• Asumsikan n = k, maka n = k+1 bernilai benar juga.

1+4+7+...+(3n-2) = n(3n-1)/2

1+4+7+...+(3k-2) = k(3k-1)/2

(1+4+7+...+(3k-2)) + 3(k+1) - 2 = (k+1)(3(k+1)-1)/2

k(3k-1)/2 + 3k + 3 - 2 = (k+1)(3k + 3 - 1)/2

(3k² - k)/2 + 3k + 1 = (k+1)(3k + 2)/2

(3k² - k)/2 + (6k + 2)/2 = (3k² + 2k + 3k + 2)/2

(3k² + 5k + 2)/2 = (3k² + 5k + 2)/2 Benar