Question

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

Answer 1

Answer:

343

Step-by-step 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 =$3(\frac{14*15}{2} )+2(14)\\=343$

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