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Hi friends,

Maybe just one more after this for today, it's 9h30 at night, and I'm getting tired....Just to mention, I do not see your help as taking advantage of your kindness and knowledge, even though it may seem so, but you are really teaching me, I study what I learn from you, and I do not take your help for granted...please believe me when I say that I really try to do these sums, many I get right, but the few ones I really cannot, I at least try them and I only ask for help here if I honestly get stuck...I'm starting to feel guilty because of all my questions...if tyou would, please assist with the following problem?

 

Prove that \(\sum_{k=3}^n (2k-1)n = n^3-4n\)

So I calculated T1=5n, T2=7n...d=2n

 

Using Tn=a+d(n-1)

I later get \(Tn=2n^2-3n\)

 

Where to from here?...Thank you for your time..

 Mar 15, 2023
 #1
avatar+1124 
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Hi friends,

 

I tried this one again this morning from a different angle..still struggle with it, I promise after this one there are no more of these...please help..

 Mar 16, 2023
 #2
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+1

 

To prove the formula for the sum of an arithmetic series, you can use the formula: 

Sn = n/2(2a + (n-1)d)

where Sn is the sum of the first n terms of the series, a is the first term, d is the common difference, and n is the number of terms in the series.

Using the values you calculated, we have: 

T1 = 5n
T2 = 7n
d = T2 - T1 = 7n - 5n = 2n

Substituting these values into the formula for the sum, we get: 

Sn = n/2(2a + (n-1)d)
   = n/2(2(5n) + (n-1)(2n))
   = n/2(10n + 2n^2 - 2n)
   = n/2(2n^2 + 8n)
   = n^2 + 4n

Therefore, the formula for the sum of the first n terms of the series is Sn = n^2 + 4n.

I hope this helps! Let me know if you have any further questions.

 Mar 16, 2023

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