Question

Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

Answer 1

Do you know the formula ?

Answer 2

What shall I say .. Erm, I do, but this one's a bit different..
the differences of the terms of this series are in AP .. :/

Answer 3

Do you know that

\[a _{n}=a _{1}+(n-1)d\]

Answer 4

obviously I do :P

Answer 5

Tell me what is a1 and what is d ?

Answer 6

This series isn't an AP , no?
I can't just take the first term and the c.d .. right?

Answer 7

my bad :(

Answer 8

right its not an AP - though the differences are an AP

Answer 9

@cwrw238 How do we change it into an AP then ?

Answer 10

I do not really remember :(

Answer 11

@cwrw238 - Exactly!

@hba - Nevermind :)

Answer 12

good question hba - i must admit i've forgotten the method - i'll have to revise!!

Answer 13

I don't even know the method :/

Answer 14

Same here :(

Answer 15

i did it in GCSE level (aged 15) - darn it
i remember a square term is involved

Answer 16

Look it up,lets do some revision :D

Answer 17

yea!!
the nth term of this sequence is n^2 + n + 1 - i think

Answer 18

n^2+n-1 i guess lol

Answer 19

n=1 1 + 1 + 1 = 3
2 2^2 + 2 + 1 = 7
3 3^2 + 3 + 1 = 13

Answer 20

yea - i guessed right!

Answer 21

now we need the sum

Answer 22

ok good :)

Answer 23

SIGMA n = n/2 [ n + 1]
SIGMA 1 = n
formula for sum of n^2 - we'd have to google that

- there might be a better way to do this - lol

Answer 24

google has it

(1/6) n(n+1)(2n+1)

Answer 25

so finally - its
sum of n terms = (1/6) n(n+1)(2n+1) + (1/2)( n + 1) + n

which can be simplified but i'm too tired!!

Answer 26

method to find nth term when difference is in AP:

seq. no. n 1 2 3 4
n^2 1 4 9 16
n^2 + n 2 6 12 20

and as the sequence is 3,7,13

nth term is n^2 + n + 1

Answer 27

method to find nth term when difference is in AP:

seq. no. n 1 2 3 4
n^2 1 4 9 16
n^2 + n 2 6 12 20

and as the sequence is 3,7,13

nth term is n^2 + n + 1

Answer 28

- i remembered it at last