Jumlah kuadrat n bilangan asli pertama 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + … n ^ 2 = 1/2n (n+1)(2n+1)​

Jawaban:

r²+3²+…+n² = n(n+1)(2n+1)/6

n = 1 (benar)

n= k –> 1²+..+k² = k(k+1)(2k+1)/6

n= k+1 –> k² + (k+1)² = (k+1)(k+1+1)(2(k+1)+1) / 6

k(k+1)(2k+1)/6 +(k+1)² = (k+1)(k+2)(2k+2+1)/6

1/6 {k(k+1)(2k+1) + 6(k+1)²} = 1/6 (k+1)(k+2)(2k+3)

1/6 {(k+1){ k(2k+1) + 6(k+1)} = 1/6(k+1)(k+2)(2k+3)

1/6 {(k+1) { 2k²+k + 6k + 6}} = 1/6(k+1)(k+2)(2k+3)

1/6 {(k+1) (2k² + 7k + 6)} = 1/6 k+1)(k+2)(2k+3)

1/6 {(k+1)(k+2)(2k +3)} = 1/6 (k+1)(k+2)(2k+3)

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