Sifat Sifat Notasi Sigma

martha yunanda baris dan deret, notasi sigma

Melanjutkan pembahasan mengenai Notasi Sigma. Sekarang akan diuraikan mengenai sifat sifat notasi sigma. Sifat ini akan mempermudah dalam penyelesaian soal notasi sigma. Berikut sifat sifat notasi sigma.

1). $\displaystyle \sum_{k=1}^{n} \, c = n . c \,$ , dengan $c \,$ adalah konstanta.

bentuk lebih umumnya : $\displaystyle \sum_{k=m}^{n} \, c = (n-m+1) . c$

2). $\displaystyle \sum_{k=m}^{n} \, c a_k = c \times \displaystyle \sum_{k=m}^{n} \, a_k$.

3). $\displaystyle \sum_{k=m}^{n} \, (a_k + b_k) = \displaystyle \sum_{k=m}^{n} \, a_k + \displaystyle \sum_{k=m}^{n} \, b_k$.

4). $\displaystyle \sum_{k=m}^{n} \, (a_k - b_k) = \displaystyle \sum_{k=m}^{n} \, a_k - \displaystyle \sum_{k=m}^{n} \, b_k$.

5). $\displaystyle \sum_{k=n}^{n} \, a_k = 0$.

6). $\displaystyle \sum_{k=m}^{n} \, a_k = \displaystyle \sum_{k=m}^{p-1} \, a_k + \displaystyle \sum_{k=p}^{n} \, a_k$.

7). $\displaystyle \sum_{k=m}^{n} \, a_k = \displaystyle \sum_{k=m+p}^{n+p} \, a_{k-p} = \displaystyle \sum_{k=m-p}^{n-p} \, a_{k+p}$.

dengan nilai $m < p < n$

Agar lebih jelas silakan lihat aplikasi sifat sifat notasi sigma tersebut dalam contoh soal dan penyelesaian di bawah ini.

1.  $\displaystyle \sum_{k=5}^{2016} \, 4$
2. $\displaystyle \sum_{k=1}^{5} \,2k$
3.  $\displaystyle \sum_{k=1}^{5} \,(k^2 + 3k)$
4.  Tentukan Hasil dari $\displaystyle \sum_{k=2}^{2016} \,(2k -1)^2 - 4\displaystyle \sum_{k=2}^{2016} \,(k^2 - k + 1) \,$ !
5.  Diketahui nilai $\displaystyle \sum_{i=1}^{36} f(i) = 245 \,$ dan $\displaystyle \sum_{i=20}^{36} f(i) = 145 , \,$ maka nilai dari $\displaystyle \sum_{i=1}^{19} f(i) \,$ adalah 
6. Bila nilai $\displaystyle \sum_{k=1}^{20} k = x \,$ . Tentukan nilai dari $\displaystyle \sum_{k=1001}^{1020} (2k - 1999) \,$ ?

Penyelesaian:

1. $\displaystyle \sum_{k=5}^{2016} \, 4$

Gunakan sifat (1) : $\displaystyle \sum_{k=m}^{n} \, c = (n-m+1) . c$

$\begin{align} \displaystyle \sum_{k=5}^{2016} \, 4 & = \underbrace{4 + 4 + 4 + ... + 4}_{\text{sebanyak } (2016 - 5 + 1) } \\ & = (2016 - 5 + 1) . 4 \\ & = (2012) . 4 \\ & = 8048 \end{align}$

2. $\displaystyle \sum_{k=1}^{5} \,2k$

Gunakan sifat (2) : $\displaystyle \sum_{k=m}^{n} \, c a_k = c \times \displaystyle \sum_{k=m}^{n} \, a_k$.

$\begin{align} \displaystyle \sum_{k=1}^{5} \,2k & = 2 \times \displaystyle \sum_{k=1}^{5} \, k \\ & = 2 \times (1 + 2 + 3 + 4 + 5) \\ & = 2 \times (15) \\ & = 30 \end{align}$

3. $\displaystyle \sum_{k=1}^{5} \,(k^2 + 3k)$

Gunakan sifat (3) : $\displaystyle \sum_{k=m}^{n} \, (a_k + b_k) = \displaystyle \sum_{k=m}^{n} \, a_k + \displaystyle \sum_{k=m}^{n} \, b_k$.

$\begin{align} \displaystyle \sum_{k=1}^{5} \,(k^2 + 3k) & = \displaystyle \sum_{k=1}^{5} \,k^2 + \displaystyle \sum_{k=1}^{5} \, 3k \\ & = (1^2 + 2^2 + 3^2 + 4^2 + 5^2) + 3 \times \displaystyle \sum_{k=1}^{5} \, k \\ & = (1 + 4 + 9 + 16 + 25) + 3 \times (1 + 2 + 3 + 4 + 5) \\ & = (55) + 3 \times (15) \\ & = (55) + 45 \\ & = 100 \end{align}$

4. $\begin{align} \displaystyle \sum_{k=2}^{2016} \,(2k -1)^2 - 4\displaystyle \sum_{k=2}^{2016} \,(k^2 - k - 3) & = \displaystyle \sum_{k=2}^{2016} \,(4k^2 - 4k + 1) - 4\displaystyle \sum_{k=2}^{2016} \,(k^2 - k + 1) \, \, \, \, \text{(sifat 2)} \\ & = \displaystyle \sum_{k=2}^{2016} \,(4k^2 - 4k + 1) - \displaystyle \sum_{k=2}^{2016} \,(4k^2 - 4k + 4) \, \, \, \, \text{(sifat 3)} \\ & = \displaystyle \sum_{k=2}^{2016} \,[ (4k^2 - 4k + 1) - (4k^2 - 4k + 4)] \\ & = \displaystyle \sum_{k=2}^{2016} \,(-3) \, \, \, \, \text{(sifat 1)} \\ & = (2016 - 2 + 1) \times (-3) \\ & = (2015) \times (-3) \\ & = -6.045 \end{align}$

5. Gunakanlah sifat (6) : $\displaystyle \sum_{k=m}^{n} \, a_k = \displaystyle \sum_{k=m}^{p-1} \, a_k + \displaystyle \sum_{k=p}^{n} \, a_k$.

$\begin{align} \displaystyle \sum_{i=1}^{36} f(i) & = \displaystyle \sum_{i=1}^{19} f(i) + \displaystyle \sum_{i=20}^{36} f(i) \\ 245 & = \displaystyle \sum_{i=1}^{19} f(i) + 145 \\ \displaystyle \sum_{i=1}^{19} f(i) & = 100 \end{align}$

Jadi, nilai $\displaystyle \sum_{i=1}^{19} f(i) = 100$

6. Gunakan sifat (7) : $\displaystyle \sum_{k=m}^{n} \, a_k = \displaystyle \sum_{k=m-p}^{n-p} \, a_{k+p}$.

$\begin{align} \displaystyle \sum_{k=1001}^{1020} (2k - 1999) & = \displaystyle \sum_{k=1001 - 1000}^{1020 -1000} (2(k+1000) - 1999) \\ & = \displaystyle \sum_{k=1}^{20} (2k + 2000 - 1999) \\  & = \displaystyle \sum_{k=1}^{20} (2k + 1) \, \, \, \, \, \text{(sifat 3)} \\ & = \displaystyle \sum_{k=1}^{20} 2k + \displaystyle \sum_{k=1}^{20} \, (1) \, \, \, \, \, \text{(sifat 1 dan 2)} \\ & = 2\displaystyle \sum_{k=1}^{20} k + (20-1+1) \times 1 \\ & = 2 x + 20 \end{align}$

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