Bank Soal Matematika SMA Notasi Sigma

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Lihat Pembahasan

Nilai dari $\sum_{i=4}^{41}\left(4i-3\right)$ adalah....

A

3.360

B

3.306

C

3.603

D

3.036

E

6.306

Pembahasan:

menggunakan sifat operasi sumasi

$\sum_{i=1}^nc=cn$
$\sum_{i=1}^n\left(a_i\pm b_i\right)=\sum_{i=1}^na_i\pm\sum_{i=1}^nb_i$
$\sum_{i=1}^nca_i=c\sum_{i=1}^na_{i\ }$
jika $m$ bilangan bulat dengan $1<m<n$ dan $a_i$ menyatakan rumus suatu barisan bilangan, maka berlaku $\sum_{i=m+1}^na_i=\sum_{i=1}^na_{i\ }-\sum_{i=1}^ma_i$
$\sum_{i=1}^ni=1+2+3+...+n=\frac{n\left(n+1\right)}{2}$

maka

$\sum_{i=4}^{41}\left(4i-3\right)=\sum_{i=4}^{41}4i-\sum_{i=4}^{41}3$ (sifat 2)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\sum_{i=4}^{41}i-\sum_{i=4}^{41}3$ (sifat 3)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\sum_{i=1}^{41}i-\sum_{i=1}^3i\right)-\left(\sum_{i=1}^{41}3-\sum_{i=1}^33\right)$ (sifat 4)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(41+1\right)}{2}-\frac{3\left(3+1\right)}{2}\right)-\left(\sum_{i=1}^{41}3-\sum_{i=1}^33\right)$ (sifat 5)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(41+1\right)}{2}-\frac{3\left(3+1\right)}{2}\right)-\left(3\left(41\right)-3\left(3\right)\right)$ (sifat 1)