Question

Find the n th term and sum of n terms of the series 5 + 8 + 17 + 44 ..........

Answer 1

it is not Ap lol

Answer 2

\(a_n = a_{n - 1} + 3^{n - 1}\)

Answer 3

Let me check if that is correct.

Answer 4

geometric progression??

Answer 5

mean

Answer 6

Yep. The formula I posted is correct for the \(n\)th term.

Answer 7

@ParthKohli wat did u did!!

Answer 8

I just tried to see the difference.
8 - 5 = 3^1
17 - 8 = 3^2
44 - 17 = 3^3

Answer 9

Nice..

Answer 10

@waterineyes Website?

Answer 11

so the 1 st order differrence is in gp

Answer 12

then...

Answer 13

Which website ?? @ParthKohli
Ha ha ha..

Answer 14

Use sigma to express the summation.

Answer 15

i also asked one of the question like this i.e\[1^2+2^2....n^2\]n=?

Answer 16

The answer to @jiteshmeghwal9's question in sigma notation is: \[\sum_{i = 1}^{n} {i^2} \] Do you want to know how I did that?

Answer 17

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

Answer 18

How did you get that? @Yahoo!

Answer 19

it is the general solution

Answer 20

I see.

Answer 21

n^3 - (n(n+1))/2)^2

Answer 22

@mukushla Can you help us?

Answer 23

ya! @ParthKohli i will be interested to know:)

Answer 24

@ParthKohli
y got it already

Answer 25

Trying to find the sum.

Answer 26

@Yahoo! has gt the right answer to my question:)

Answer 27

@ParthKohli the answer given in my book is matching with @Yahoo! 's answer:)

Answer 28

i have determine a pattern a pattern of @Yahoo! 's question

Answer 29

@jiteshmeghwal9 can u show that

Answer 30

\(5 + 3 \Longrightarrow 2^{nd}\)
\(5 + 3 + 3^2 \Longrightarrow 3^{rd}\)

I am not sure how to represent these in sigma notation.

Answer 31

\(a_n=a_{n-1}+3^{n-1}=a_{n-2}+3^{n-1}+3^{n-2}=...=a_1+3+...+3^{n-2}+3^{n-1} \)

Answer 32

Any other way to represent that?

Answer 33

wht @ParthKohli replied after ur reply @Yahoo!