3. Diketahui pola 1² +2²+3²+ …… + n²

1² + 2² + 3² + 4² + 5² + … + n²

= 1 + 4 + 9 + 16 + 25 + … + n²

<=> 1, 5, 14, 30, 55, … ,

1 (a)

5 – 1 = 4 (b)

14 – 5 = 9

30 – 14 = 16

55 – 30 = 25

9 – 4 = 5 (c)

16 – 9 = 7

25 – 16 = 9

7 – 5 = 2 (d)

9 – 7 = 2

Un=a+(n-1)b+(n-1)(n-2)c:2+(n-1)(n-2)(n-3)d:6

Un=1+(n-1)4+(n-1)(n-2)5:2+(n-1)(n-2)(n-3)2:6

6Un=6+(n-1)24+(n-1)(n-2)15+(n-1)(n-2)(n-3)2

6Un=6+(n-1)[24+(n-2)15+(n²-5n+6)2]

6Un=6+(n-1)(24+15n-30+2n²-10n+12)

6Un=6+(n-1)(24-30+12+15n-10n+2n²)

6Un=6+(n-1)(6+5n+2n²)

6Un=6+(n-1)(2n²+5n+6)

6Un=6+2n³+5n²+6n-2n²-5n-6

6Un=2n³+3n²+n

6Un=n(2n²+3n+1)

6Un=n(2n²+2n+n+1)

6Un=n(2n(n+1)+(n+1))

6Un=n(n+1)(2n+1)

Un= [n(n+1)(2n+1)]:6

Sehingga Jumlah n suku dari deret

1²+2²+3²+…+n²

= Sn

= [n(n+1)(2n+1)]:6

semoga bisa sedikit membantu, maaf tidak hafal rumusnya, hingga terpaksa dicari dahulu.