Sum the series 1x3+3x6+5x9+ ... up to n terms. Further I want to know how do we find the common difference of such a sequence (d) in order to find term k= a +(k-1)d. And the use term K to find the sum.

Question

ganeshie8 · Answer

start by finding the general term

ganeshie8 · Answer

the $n^{th}$ term..

anonymous · Answer

how do we find k that is the difference first the book says k is 2

ganeshie8 · Answer

what is "k" ?

anonymous · Answer

oh sorry my bad d is 2 what I get is that they did 3-1=2 or 5-3=2 to get d but how dowe know which no of the sequence to subtract for common difference if we use 6-3 it gives us 3

ganeshie8 · Answer

Got you :) 

I would start like this :-

I notice that each term has two parts, so the $n^{th}$ term need to look somethig like  :     $part1 \times     part2$

anonymous · Answer

hmmm go on

ganeshie8 · Answer

$part2$  is easy to figure out :   its simply  $3n$,

so the $n^{th}$ term becomes :  $part1\times   3n$

ganeshie8 · Answer

does this really look correct ?
lets check it once quick

ganeshie8 · Answer

$a_n =   part1 \times 3n$

put n = 1, you wud get   :   $a_1 =  part1 \times  3$

put n = 2, you wud get   :   $a_2 =  part1 \times  6$

put n = 3, you wud get   :   $a_3 =  part1 \times  9$

...

ganeshie8 · Answer

looks like we're right, it has to be $3n$
so lets go ahead and figure out $part1$

anonymous · Answer

ok

ganeshie8 · Answer

$part1$  has below pattern :

1, 3, 5, .... 

right ?

anonymous · Answer

yes

ganeshie8 · Answer

does that look like a familiar sequence ?

anonymous · Answer

I have A query how do we get to know whether we have to take part 1 or part 2 to take out d?

ganeshie8 · Answer

for both parts we have a "d",
but there is no "d" for the series as a whole

ganeshie8 · Answer

the "d" for first part is "3"
the "d" for second part is "2"

anonymous · Answer

ssecond is 3 first is 2

ganeshie8 · Answer

good catch :) you're right !!

anonymous · Answer

;D but which one are we supposed to take?

ganeshie8 · Answer

wat do we get for $part1$ ?

ganeshie8 · Answer

there is NO "d" for the series as whole,
lets keep going :)

anonymous · Answer

Okay I think I got it if we take d is 2 then we can take first term as 1 , while if we take d as 3 we can take first term as 3.

anonymous · Answer

we form an equation and get term k (if we take d as 2) 2k-1 while if we take d as 3 Term k is: 3k

ganeshie8 · Answer

yes :)


for $part1$ :

1, 3, 5...


since this is an arithmetic sequence,
$n^{th}$ term,  $a_n = 1 +  (n-1)\times 2  = 1 + 2n -2  = 2n -1 $

ganeshie8 · Answer

so, the $n^{th}$ term of series is :

$a_n =      (2n-1)(3n)$

anonymous · Answer

we multiply both these terms to get the final term k that is 2k-1 * 3k = 6k^2 -3k

ganeshie8 · Answer

looks good

anonymous · Answer

well thank you I know the rest as to how to sum it if it is k ,k^2,k^3 etc

ganeshie8 · Answer

Sum of first $n$ terms =   $\large \sum \limits_{k=1}^n (6k^2 - 3k)$

ganeshie8 · Answer

yes, rest should be easy... have fun ;P

anonymous · Answer

yup