What is the algebraic expression for sum of the sequence 7,11,157,11,15?

2 Answers
Feb 19, 2018

2n^2+5n2n2+5n

Explanation:

The sum of the sequence means adding;

7+11=187+11=18

18+15=3318+15=33

This means the sequence turns to 7,18,337,18,33

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

33-18=153318=15

18-7=11187=11

Finding the difference of the differences:

15-11=41511=4

To find the quadratic of the N'th term, we divide this by 22, giving us 2n^22n2

Now we take away 2n^22n2 from the original sequence:

1n^2=1,4,9,16,25,361n2=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

Feb 19, 2018

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 )