Evaluate the summation of 3 n plus 2, from n equals 1 to 14..

We have to evaluate the summation of (3n+2), with n ranging from 1 to 14. Let's write all the terms of the sum for each value of n:
n=1: [tex] 3n+2 = 3\cdot 1+2=5[/tex]
n=2: [tex]3n+2 = 3\cdot 2+2=8[/tex]
n=3: [tex]3n+2 = 3\cdot 3+2=11[/tex]
n=4: [tex]3n+2 = 3\cdot 4+2=14[/tex]
n=5: [tex]3n+2 = 3\cdot 5+2=17[/tex]
n=6: [tex]3n+2 = 3\cdot 6+2=20[/tex]
n=7: [tex]3n+2 = 3\cdot 7+2=23[/tex]
n=8: [tex]3n+2 = 3\cdot 8+2=26[/tex]
n=9: [tex]3n+2 = 3\cdot 9+2=29[/tex]
n=10: [tex]3n+2 = 3\cdot 10+2=32[/tex]
n=11: [tex]3n+2 = 3\cdot 11+2=35[/tex]
n=12: [tex]3n+2 = 3\cdot 12+2=38[/tex]
n=13: [tex]3n+2 = 3\cdot 13+2=41[/tex]
n=14: [tex] 3n+2 = 3\cdot 14+2=44[/tex]
Now, let's sum all the terms together, and we get:
[tex]S=5+8+11+14+17+20+23+26+29+32+35+38+41+44=343[/tex]