Question

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


please help

Answer 1

Answer:

We are given a problem involving summation .

$\sum_{n=1}^{14} 3n+2$

$\implies 3(1)+2+3(2)+2+.......3(14)+2\\\\\implies 3+2+6+2+....42+2\\\\\implies (3+6+9.....42)+(2\times 14)\\\\\implies \dfrac{n}{2}[2a+(n-1)d]+28\\\\\implies \dfrac{14}{2}[2(3)+(14-1)(3)]+28\\\\\implies 7[6+13\times 3]+28\\\\\implies 7[6+39]+28\\\\\implies 7\times 45 + 28\\\\\implies 315+28\\\\\implies 343$

∴ The value will be 343 .

Step-by-step explanation:

The only formula involved here is the calculation of the sum of an arithmetic progression ( A.P ) .

Let the first term be a .

Let the common difference be d .

Then the sum of n terms is given by the formula :

$\dfrac{n}{2}[2a+(n-1)d]$

Summation starts from an initial value and ends in the above value .

Answer 2

Solution :

Sigma ( 14 => 3n +2)( n = 1)

➨ 3(1) + 2 + 3(2) + 2+........ 3(14) +2

➨ 3 + 2 + 6 + 2 + ..... 42 + 2

➨ (3 + 6 + 9 ......42) + ( 2 × 14 )

➨ n/2[ 2a + ( n - 1)d] + 28

➨ 14/2 [ 2(3) + (14 - 1)(3)] + 28

➨ 7( 6 + 13 × 3 ) + 28

➨ 7( 6 + 39 ) + 28

➨ 7 × 45 + 28

➨ 315 + 28

➨ 343

Hence ,

.The Value will be 343