WebSep 15, 2024 · Find each sum. (3n2 – 5n + 6) + (–8n2 – 3n – 2) = Get the answers you need, now! sumanriddam9821 sumanriddam9821 16.09.2024 ... Secondary School … WebDec 5, 2014 · Here you can see that we can assume the sum of the numbers up through $3n-2$ is $\frac{n(3n-1)}{2}$, and this fact is used in the very first equation. The rest of the work is to show that when you add the next term ($3(n+1) - 2$) to this, are able to show the fact that is to be proved, which is that the sum of terms through $3(n+1) - 2$ is ...
Solved Determine whether the sequence converges or diverges
Webb. c. d. e. f. Find each sum (3n2-5n+6)+(-8n2-3n-2) Find the midpoint of the segment with the following endpoints. WebMar 3, 2013 · Your recursive equation is pretty simple: every time you expand T(n/2) only one new term appears. So in the non-recursive form you'll have a sum of quadratic terms. You only have to show that A(n) is log(n) (I leave it to you as an easy exercise). explorer pop up tent
Solve 3n^2-n-10 Microsoft Math Solver
WebFind each sum.(3n2 - 5n+6) + (-8n2 – 3n - 2) =-11n2 - 8n-4-11n2 - 2n + 4-5n2 - 8n + 4-512 – 2n - 4. Answer by Guest (3n2-5n+6)+(-8n2-3n-2) 3n2-5n+6-8n2-3n-2 Collect like … WebJul 6, 2013 · In this case compare all the terms along with their coefficients with leading term and replace the leading term in all other terms without coefficient then sum up the term to get a single term with highest coefficient and that coefficient is treated as c. in second case take the least coefficient as k which should be greater than x and compare ... WebJan 18, 2024 · Let the sum of n terms be given by Sn . so. Sn = 3n²/2+ 5n/2. S1 = 3(1)²/2 + 5(1)/2 = 3/2+5/2 => 4 . so 1st term is 4 say 'a' Now . S2 = 3(2)²/2 + 5(2)/2 = 6+5 => 11 . Now a2 = S2 - a1 => a2 = 11-4 = 7 . Now common difference (d) = a2-a1 => 7-4 = 3 . we know . an = a+(n-1) d. so . a25 = 4 + (25-1)(3) => a25 = 4 + 24×3 =>4+ 72. so 25th term ... explorer range backpack