# Soal sigma 1B
###### tags: `math` `sigma`
## Soal
$$
sederhanakan \sum_{a=2}^{10}({a}^{3}+1)+\sum_{a=2}^{10}({a}^{2}+1)
$$

## Penyelesaian
dari sifat
$$
\sum_{i=1+p}^{n+p}(U_{i-p})
$$
maka :
$$
\sum_{a=2}^{10}({a}^{3}+1) = \sum_{a=2}^{10}U_{a} = \sum_{a=1+1}^{9+1}(U_{a-1})
$$

$$
U_{a}={a}^{3}+1
$$

$$
U_{a-1}={(a-1)}^{3}+1
$$

berarti :
$$
\sum_{a=2}^{10}({a}^{3}+1)=\sum_{a=1+1}^{9+1}U_{a-1}=\sum_{a=1}^{9}U_{a-1}=\sum_{a=1}^{9}({(a-1)}^{3}+1)
$$
begitu juga yg sigma satunya
$$
\sum_{a=2}^{10}({a}^{2}+1) = \sum_{a=2}^{10}U_{a} = \sum_{a=1+1}^{9+1}(U_{a-1})
$$

$$
U_{a}={a}^{2}+1
$$

$$
U_{a-1}={(a-1)}^{2}+1
$$

berarti :
$$
\sum_{a=2}^{10}({a}^{2}+1)=\sum_{a=1+1}^{9+1}U_{a-1}=\sum_{a=1}^{9}U_{a-1}=\sum_{a=1}^{9}({(a-1)}^{2}+1)
$$

dari sifat :
$$
\sum_{i=1}^{n}(U_i+V_i)=\sum_{i=1}^{n}U_i+\sum_{i=1}^{n}V_i
$$
maka:
$$
\sum_{a=2}^{10}({a}^{3}+1)+\sum_{a=2}^{10}({a}^{2}+1)
$$

$$
=\sum_{a=1}^{9}({(a-1)}^{3}+1)+\sum_{a=1}^{9}({(a-1)}^{2}+1)
$$

$$
=\sum_{a=1}^{9}({(a-1)}^{3}+1+{(a-1)}^{2}+1)
$$

$$
=\sum_{a=1}^{9}({(a-1)}^{3}+{(a-1)}^{2}+2)
$$

## Jawaban
jadi bentuk sederhananya adalah :
$$
\sum_{a=2}^{10}({a}^{3}+1)+\sum_{a=2}^{10}({a}^{2}+1)=\sum_{a=1}^{9}({(a-1)}^{3}+{(a-1)}^{2}+2)
$$