MHB Summing Formula Problem [Again]

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The discussion revolves around creating sum formulas for two arithmetic series: the first series is 1 + 3 + 5 + ... + 99, and the second is 2 + 6 + 10 + ... + 106. The formula for the sum of an arithmetic series is provided, which is S(n) = n/2 [2a + (n-1)d], where 'a' is the first term, 'd' is the common difference, and 'n' is the number of terms. For the second series, the values of a, d, and n are identified as a=2, d=4, and n=26, leading to a calculated sum of 1352. The discussion encourages users to show their progress to receive better assistance.
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hello expert..

how do make sum formula for this below problem (2 question) :

please, see my attachment picture

thanks in advance..

susanto3311
 

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Hello, susanto3311![

How to make formulas for these sums?

$[1]\;1+3+5+7\cdots+99$

$[2]\;2+6+10+14+\cdots+106$
These two happen to be arithmetic series.

The sum of the first $n$ terms of an arithmetic series
$\quad$is given by: $\:S(n) \;=\;\frac{n}{2}\big[2a + (n-1)d\big]$

$\qquad\text{where: }\:\begin{Bmatrix} a &=& \text{first term} \\ d &=& \text{common difference} \\ n&=& \text{number of terms} \end{Bmatrix}$In [2], we have: $\:a=2,\;d=4,\;n=26$

Therefore: $\:S \:=\:\frac{26}{2}\big[2(2) + 25(4)\big]\:=\:1352$
 
We do ask that people posting questions show their progress so that our helpers know where you are stuck and can offer the best help possible. :D

The sum of the members of a finite arithmetic progression is called an arithmetic series, and is given by:

$$S_n=\frac{n}{2}\left(a_1+a_n\right)$$

To determine $n$, I would think of the terms of the first progression as $2n-1$ and for the second $2(2n-1)$ (and so how will the two sums be related?). Can you proceed?
 
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