Question

Find the sum upto n terms of the below series,
3.8 + 6.11 + 9.14 + ...

Here ' . ' between two numbers is used for product not for decimal.​

Answer 1

$\large \dag$ Question :-

Find the sum upto n terms of the below series,

3.8 + 6.11 + 9.14 + ...

Here ' . ' between two numbers is used for product not for decimal.

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{The   \: Sum \:  is  \: 3n(n+1)(n+3) }} }$

$\large \dag$ Step by step Explanation :-

The given series is product of two arithmetic series.

We know that nth term of product of two arthematic series is given by product nth term of each arthematic series.

☆ Taking 1st Series :- 3 , 6 , 9 , . . . .

$\\ \text T_{n_{_1}} = 3 + (n - 1) 3$

$:\longmapsto \text T_n = 3 + 3n - 3$

$\large:\longmapsto \underline{ \underline{ \text T_n = 3n}}$

☆ Taking 2nd Series :- 8 , 11 , 14 , . . . .

$\\ \text T_{n_{_2}}= 8 + (n - 1) 3$

$:\longmapsto \text T_n = 8 + 3n - 3$

$\large:\longmapsto \underline{ \underline{ \text T_n = 3n+5}}$

Therefore,

nth term of resulting arithmetic series is ;

$\\\text T_n = \text T_{n_{_1}} . \: \text T_{n_{_2}}$

$:\longmapsto \text T_n = 3n(3n + 5)$

$\purple{ \large :\longmapsto  \underline {\boxed{{ \pmb{\text T_n = 9 {n}^{2} + 15n } }}}}$

Now let the sum of given series upto n terms be $\sf S_n$ ;

$\\:\longmapsto \text S_n = \sum \limits^n_{k = 1} a_k$

$= \sum \limits^n_{k = 1}(9 {k}^{2} + 15k)$

$= 9\sum \limits^n_{k = 1} {k}^{2} + 15\sum \limits^n_{k = 1}k$

$= 9 \times \frac{n(n + 1)(2n + 1)}{6} + 15 \times \frac{n(n + 1)}{2}$

$= \frac{3n(n + 1)(2n + 1)}{2} + \frac{15n(n + 1)}{2}$

$= \frac{3n(n + 1)(2n + 6)}{2}$

$= \frac{3n(n + 1)2 \times (n + 3)}{2}$

$\blue{ \large :\longmapsto  \underline {\boxed{{\bf S_n = 3n(n + 1)(n + 3)} }}}$

Answer 2

Answer:

$\large \dag$ Answer :-

• The sum upto n terms of the series

• 3.8 + 6.11 + 9.14 +_____

• Is ,____?

$\large \dag$ Explanation for your question. :-

• Hey mate ,

• Here kindly refer the attachment for more information.

• And for good understanding.

$\large \dag$ Some Hints :-

• Here first you should come to know that n terms of the product of given arithmetic series .

• Taking now by 1st series as :-3,6,9,____.

• Next,

• Taking second series as :-8,11,14,____.

• And ,at final we will get nth terms resulting arithmetic series (By multiplying Tn=Tn1 × Tn2 we get the nth terms ).

• Next ,by taking sum series upto nth terms .

• Then,we will get total sum .

Therefore,

• This is the perfect answer to your question..