[tex]\small\boxed{\begin{aligned}\sf \sum\limits_{k = 1}^{5} \: (2a + k)^2&=\sf \sum\limits_{k = 1}^{5} \: ( {(2a)}^{2} + 2(2ak) + {k}^{2} )\\ \sf &=\sf \sum\limits_{k = 1}^{5} \: ( {4a}^{2} + 4ak + {k}^{2} )\\ \sf &= \sf \sum\limits_{k = 1}^{5} \: {4a}^{2} + \sum\limits_{k = 1}^{5} \: 4ak + \sum\limits_{k = 1}^{5} \: {k}^{2} \\ \sf &=\sf\\ \sf &= \sf {4a}^{2}(5 - 1 + 1) + 4a(1 + 2 + 3 + 4 + 5) + ( {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} + {5}^{2} ) \\ \sf &=\sf {4a}^{2}(5) + 4a(15) + (1 + 4 + 9 + 16 + 25)\\ \sf &= \sf {20a}^{2} + 60a + 55 \end{aligned}}[/tex]
NOtasi SIgma
penjumlahan
Penjelasan dengan langkah-langkah:
bentuk sederhana dari
[tex]\sf \sum\limits_{k = 1}^{5}\ (2a + k)^2=[/tex]
[tex]\sf=\sum\limits_{k = 1}^{5}\ (4a^2 + 4ak + k^2)\\\\[/tex]
[tex]\sf=\sum\limits_{k = 1}^{5}\ (k^2 + 4ak + 4a^2)\\\\[/tex]
[tex]\sf=\sum\limits_{k = 1}^{5}\ k^2 + \sum\limits_{k = 1}^{5} 4ak + \sum\limits_{k = 1}^{5}\ 4a^2[/tex]
[tex]\sf=\sum\limits_{k = 1}^{5}\ k^2 +4a \sum\limits_{k = 1}^{5} k + 4a^2\sum\limits_{k = 1}^{5}\ 1[/tex]
[tex]\sf =(1^2 +2^2 + 3^2+4^2+5^2) + 4a (1 +2 +3+4+ 5) + 4a^2(1)(5)[/tex]
= 55 + 4a(15) + 20a²
= 20a² + 60 a + 55
[answer.2.content]
