Bank Soal Matematika SMA Notasi Sigma

Perhatikan beberapa sifat notasi sigma berikut!

1. $\sum_{i=1}^nU_i=U_1+U_2+U_3+\dots+U_n$
2. $\sum_{i=1}^nU_i=\sum_{j=1}^nU_j$
3. $\sum_{i=1}^nC=Cn$
4. $\sum_{i=1}^nCU_i=C\sum_{i=1}^nU_i$
5. $\sum_{i=1}^n\left(U_i\pm V_i\right)=\sum_{i=1}^nU_i\pm\sum_{i=1}^nV_i$
6. $\sum_{i=1}^n\left(U_i+V_i\right)^2=\sum_{i=1}^nU_i^2+2\sum_{i=1}^nU_iV_i+\sum_{i=1}^nV_i^2$
7. $\sum_{i=1}^nU_i+\sum_{i=n+1}^mU_i=\sum_{i=1}^mU_i$
8. $\sum_{i=1}^nU_i=\sum_{i=0}^{n-1}U_{i+1}=\sum_{i=2}^{n+1}U_{i-1}$

dengan

$n$ dan $m$: suatu bilangan bulat non-negatif

$U_i$: rumus suku ke-$i$ suatu deret

$V_i$: rumus suku ke-$i$ suatu deret

$C$: suatu konstanta

Diketahui notasi sigma berikut

$\sum_{n=1}^{10}(n-4)^2$

Akan dicek untuk masing-masing pilihan jawaban.

Pilihan jawaban $\sum_{n=1}^{10}\left(n^2-8n+16\right)$ sama dengan notasi sigma $\sum_{n=1}^{10}(n-4)^2$ sebab

$n^2-8n+16=\left(n-4\right)^2$

Pilihan jawaban $\sum_{n=1}^{10}n^2-8\sum_{n=1}^{10}n+160$ sama dengan notasi sigma $\sum_{n=1}^{10}(n-4)^2$ sebab

berdasarkan sifat 6 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{n=1}^{10}n^2+2\sum_{n=1}^{10}n\left(-4\right)+\sum_{n=1}^{10}\left(-4\right)^2$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{n=1}^{10}n^2+2\sum_{n=1}^{10}\left(-4n\right)+\sum_{n=1}^{10}16$

berdasarkan sifat 4 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{n=1}^{10}n^2+2\left(-4\right)\sum_{n=1}^{10}n+\sum_{n=1}^{10}16$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{n=1}^{10}n^2-8\sum_{n=1}^{10}n+\sum_{n=1}^{10}16$

berdasarkan sifat 3 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{n=1}^{10}n^2-8\sum_{n=1}^{10}n+16.10$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{n=1}^{10}n^2-8\sum_{n=1}^{10}n+160$

Pilihan jawaban $\sum_{i=1}^3\left(i-4\right)^2+\sum_{i=4}^{10}\left(i-4\right)^2$ sama dengan notasi sigma $\sum_{n=1}^{10}(n-4)^2$ sebab

berdasarkan sifat 2 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{i=1}^{10}\left(i-4\right)^2$

berdasarkan sifat 7 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{i=1}^3\left(i-4\right)^2+\sum_{i=4}^{10}\left(i-4\right)^2$

Pilihan jawaban $\sum_{k=0}^9\left(k-3\right)^2$ sama dengan notasi sigma $\sum_{n=1}^{10}(n-4)^2$ sebab

berdasarkan sifat 2 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{k=1}^{10}\left(k-4\right)^2$

berdasarkan sifat 8 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{k=0}^{10-1}\left(\left(k+1\right)-4\right)^2$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{k=0}^9\left(k+1-4\right)^2$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{k=0}^9\left(k-3\right)^2$

Pilihan jawaban $\sum_{j=2}^{11}\left(j-4\right)^2$ tidak sama dengan notasi sigma $\sum_{n=1}^{10}(n-4)^2$ sebab

berdasarkan sifat 2 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{j=1}^{10}\left(j-4\right)^2$

berdasarkan sifat 8 diperoleh

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{j=2}^{10+1}\left(\left(j-1\right)-4\right)^2$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{j=2}^{11}\left(j-1-4\right)^2$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{j=2}^{11}\left(j-5\right)^2$

$\sum_{n=1}^{10}\left(n-4\right)^2=\sum_{j=2}^{11}\left(j-5\right)^2\ne\sum_{j=2}^{11}\left(j-4\right)^2$