# Chapter 11 - Section 11.1 - Sequences and Summation Notation - Exercise Set - Page 829: 44

$\sum_{i=1}^{14}(2i+4)$

#### Work Step by Step

Since, we have $6+8+10+12+.....32$ This form an arithmetic progression sequence whose first term is $6$ and last term is $32$ with common difference of $2$. For an arithmetic progression sequence, we have $a_n=a_1+(n-1)d$ thus, $a_n=6+(n-1)2=6+2n-2=2n+4$ when $n=14$ Lat term becomes $a_{14}=2(14)+4=32$ Hence, $6+8+10+12+.....32$ can be summed up with the summation notation as: $\sum_{i=1}^{14}(2i+4)$

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