Question

sum to n terms of 3.5+4.7+5.9+.....

Answer 1

hope it helps
good luck

Answer 2

Answer:

$S_{n}=\frac{2(n)(n+1)(2n+1)}{6} +\frac{7(n)(n+1)}{2}+6n$

Step-by-step explanation:

The first term is $p_{1}=3.5=(1+2)(2.1+3)$

The second term is $p_{2}=4.7=(2+2)(2.2+3)$

The third term is $p_{3}=5.9=(3+2)(2.3+3)$

Following the same pattern, the nth term will be $p_{n}=(n+2)(2n+3)$

$p_{n}=2n^{2} +3n+4n+6$

$p_{n}=2n^{2} +7n+6$

Therefore, the sum of n terms will be: $S_{n}={\sum}t_{n}={\sum}(2n^{2}+7n+6)=2{\sum}n^{2}+7{\sum}n+6{\sum}1$

$S_{n}=\frac{2(n)(n+1)(2n+1)}{6} +\frac{7(n)(n+1)}{2}+6n$