Question

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

Answer 1

343

Explanation:

WE are to evaluate the sum of

`3n+2` from n=1 to 14

Symbolically this can be reprsented as

`S=\Sigma _(n=1)  ^(14) (3n+2)`

Splitting this into two separate terms we have

`S=\Sigma _(n=1)  ^(14) 3n+\Sigma _(n=1)  ^(14) 2`

Since in I term 3 is a constant 3 can be taken out and similarly for II term if 2 taken out it becomes sum of 1, 14 times

`S=3\Sigma _(n=1)  ^(14) n+    28\\`

The first term is sum of 14 natural numbers and we use the formula

Sum =
`3((14*15)/(2) )+2(14)\\=343`

Answer 2

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:
`3n+2 = 3\cdot 1+2=5`
n=2:
`3n+2 = 3\cdot 2+2=8`
n=3:
`3n+2 = 3\cdot 3+2=11`
n=4:
`3n+2 = 3\cdot 4+2=14`
n=5:
`3n+2 = 3\cdot 5+2=17`
n=6:
`3n+2 = 3\cdot 6+2=20`
n=7:
`3n+2 = 3\cdot 7+2=23`
n=8:
`3n+2 = 3\cdot 8+2=26`
n=9:
`3n+2 = 3\cdot 9+2=29`
n=10:
`3n+2 = 3\cdot 10+2=32`
n=11:
`3n+2 = 3\cdot 11+2=35`
n=12:
`3n+2 = 3\cdot 12+2=38`
n=13:
`3n+2 = 3\cdot 13+2=41`
n=14:
`3n+2 = 3\cdot 14+2=44`
Now, let's sum all the terms together, and we get:

`S=5+8+11+14+17+20+23+26+29+32+35+38+41+44=343`