# What is the algebraic expression for sum of the sequence 7,11,15?

Feb 19, 2018

$2 {n}^{2} + 5 n$

#### Explanation:

The sum of the sequence means adding;

$7 + 11 = 18$

$18 + 15 = 33$

This means the sequence turns to $7 , 18 , 33$

We want to find the N'th term, we do this by finding the difference in the sequence:

$33 - 18 = 15$

$18 - 7 = 11$

Finding the difference of the differences:

$15 - 11 = 4$

To find the quadratic of the N'th term, we divide this by $2$, giving us $2 {n}^{2}$

Now we take away $2 {n}^{2}$ from the original sequence:

$1 {n}^{2} = 1 , 4 , 9 , 16 , 25 , 36$

$\therefore$ $2 {n}^{2} = 2 , 8 , 18 , 50 , 72$

We only need the first $3$ sequences:

$7 - 2 = 5$

$18 - 8 = 10$

$33 - 18 = 15$

Finding the difference between the differences:

$15 - 10 = 5$

$10 - 5 = 5$

Therefore we $+ 5 n$

This gives us:

$2 {n}^{2} + 5 n$

We can check this by substituting the values of $1 , 2 \mathmr{and} 3$

$2 {\left(1\right)}^{2} + 5 \left(1\right) = 2 + 5 = 7$ So this works...

$2 {\left(2\right)}^{2} + 5 \left(2\right) = 8 + 10 = 18$ So this works...

$2 {\left(3\right)}^{2} + 5 \left(3\right) = 18 + 15 = 33$ So this works...

$\therefore$ the expression = $2 {n}^{2} + 5 n$

Feb 19, 2018

Alternate...

#### Explanation:

The sequence is defined by: ${a}_{n} = 4 n + 3$

Hence we are trying to find the sum of the first $n$ terms...

$7 + 11 + 15 + \ldots + 4 n + 3$

In sigma notation

$\implies {\sum}_{r = 1}^{n} 4 r + 3$

We can use our knowledge of series...

$\sum c {n}^{2} + a n + b \equiv c \sum {n}^{2} + a \sum n + b \sum 1$

We also know..

${\sum}_{r = 1}^{n} 1 = n$

${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

$\implies \sum 4 n + 3 = 4 \sum n + 3 \sum 1$

$\implies 4 \cdot \left(\frac{1}{2} n \left(n + 1\right)\right) + 3 n$

$\implies 2 n \left(n + 1\right) + 3 n$

$\implies 2 {n}^{2} + 2 n + 3 n$

$\implies 2 {n}^{2} + 5 n$

$\implies n \left(2 n + 5\right)$