Question

silahkan dicoba yaa ​

Answer 1

Jawab:

B

Penjelasan dengan langkah-langkah:

Rumus-rumus yang digunakan

[tex]\begin{aligned}&\sum_{k=1}^{n}c=cn\\&\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\&\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\end{aligned}[/tex]

[tex]\begin{aligned}&21^2+22^2+23^2+24^4+25^2+26^2\\&=\sum_{k=1}^{6}(20+k)^2\\&=\sum_{k=1}^{6}(400+40k+k^2)\\&=\sum_{k=1}^{6}400+\sum_{k=1}^{6}40k+\sum_{k=1}^{6}k^2\\&=400(6)+40~\frac{6(6+1)}{2}+\frac{6(6+1)(2\cdot 6+1)}{6}\\&=2400+840+91\\&=3331\end{aligned}[/tex]