Question

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

Answer 1

`2n^2+5n`

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 `2n^2`

Now we take away `2n^2` from the original sequence:

`1n^2=1,4,9,16,25,36`

`therefore` `2n^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 `+5n`

This gives us:

`2n^2+5n`

We can check this by substituting the values of `1, 2 and 3`

`2(1)^2+5(1)=2+5=7` So this works...

`2(2)^2+5(2)=8+10=18` So this works...

`2(3)^2+5(3)=18+15=33` So this works...

`therefore` the expression = `2n^2+5n`

Answer 2

Alternate...

Explanation:

The sequence is defined by: `a_n = 4n+3`

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

`7 + 11 + 15 + ... + 4n+3`

In sigma notation

`=> sum_(r=1) ^n 4r+3`

We can use our knowledge of series...

`sum cn^2 + an +b -= c sum n^2 + asum n + b sum 1`

We also know..

`sum_(r=1) ^n 1 = n`

`sum_(r=1) ^n r = 1/2 n (n+1)`

`=> sum 4n + 3 = 4sumn + 3sum1`

`=> 4 * ( 1/2 n (n+1 ) ) + 3n`

`=> 2n(n+1) + 3n`

`=> 2n^2 +2n + 3n`

`=> 2n^2 + 5n`

`=> n(2n+5 )`