Question

Sum the series to n terms -
1.4.7 + 4.7.10 + 7.10.13 +...........................to n terms.

Answer 1

@mukushla

Answer 2

\[a_n=(3n+1)(3n+4)(3n+7)\]

Answer 3

\[n=0,1,2,...\]

Answer 4

i think there will be (3n-2) in place of 3n+7

Answer 5

n starts from 0 ... check again :)

Answer 6

but n is natural no.

Answer 7

it doesn't matter to use natural or start from 0 ha?

Answer 8

because the result is same

Answer 9

these series are applicable only when n is natural.

Answer 10

ok we turn it to natural...\[a_n=(3n-2)(3n+1)(3n+4)\]\[n=1,2,3,...\]

Answer 11

now?

Answer 12

one way (uglyiest) is to multiply all of brackets and using \(\sum k\) , \(\sum k^2\) and \(\sum k^3\)

Answer 13

but there are nice methods...

Answer 14

??

Answer 15

guys what do u think?

Answer 16

after simplifying u have\[a_n=27n^3+27n^2-18n-8\]

Answer 17

i have solved like this-
\[a_{n} = 27n^{3}+27n^{2}-18n-8\]
\[S_{n}=\sum{a_{n}}\]
\[S_{n}=27\sum{n^{3}}+27\sum{n^{2}}-18\sum{n}-8n\]

Answer 18

exactly

Answer 19

did i interrupt mukushla? I think I did 'answer snipping' that is a hard one for me ..

sorry you may continue
though I promise I will interrupt at last

Answer 20

\[\sum k^3=(\frac{n(n+1)}{2})^2\]\[\sum k^2=\frac{n(n+1)(2n+1)}{6}\]\[\sum k=\frac{n(n+1)}{2}\]

Answer 21

i know all these values, continue.........

Answer 22

@mathslover
np ... :)

Answer 23

\[S_n=\frac{n}{4}(27n^3+90n^2+45n-50)\]this is what u will got after simplifying according to god of wolf

Answer 24

can it be simplified more

Answer 25

???

Answer 26

it must divisible by 4 so yeah i think it can...let me ask wolf

Answer 27

http://www.wolframalpha.com/input/?i=%283n-2%29%283n%2B1%29%283n%2B4%29%283n%2B7%29%2F12+%2B+56%2F12

Answer 28

it is the ans.

Answer 29

see its expanded form and match with yours.

Answer 30

so its ok

Answer 31

yes it is correct

Answer 32

how will you simplify it further?

Answer 33

The above series can be written in the form:
tn = (1+3n)^3 -9(1+3n), where n=1,2,3...

Answer 34

thanks i got it.