+1124 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..
+1124 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..
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.
