tentukan 3 suku pertama dan notasi sigmanya dari 1.4n 2. n+2 3.2-n

[tex]\displaystyle \sum^{3}_{n=1}4n=4(1)+4(2)+4(3)\\\sum^{3}_{n=1}4n=4+8+12\\\boxed{\boxed{\sum^{3}_{n=1}4n=24}}\\\\\\\sum^{3}_{n=1}n+2=(1+2)+(2+2)+(3+2)\\\sum^{3}_{n=1}n+2=3+4+5\\\boxed{\boxed{\sum^{3}_{n=1}n+2=12}}\\\\\\\sum^{3}_{n=1}2-n=(2-1)+(2-2)+(2-3)\\\sum^{3}_{n=1}2-n=1+0-1\\\boxed{\boxed{\sum^{3}_{n=1}2-n=0}}[/tex]