Find the sum \(\displaystyle \frac{1}{2} + \frac{5}{4} + \frac{14}{8} + \frac{30}{16} + \dots + \frac{1^2 + 2^2 + \dots + n^2}{2^n} + \dotsb\)

Guest May 12, 2020

#1**0 **

sumfor(n, 1, 10000, ((2 *n + 1) *(n^2 + n)/(6 *2^n) + n^2 / 2^n) **=first part =12 + 2nd part =6 =12+6 =18**

Guest May 12, 2020

#2**0 **

Notice the numerator in the first part of this sequence is: 1 + The second square(2^2) =5. The 3rd term is: 5 + The 3rd square(3^2) =14. And 14 + 4^2 =30......and so on. The "closed form formula" for these numerators is: 1/6 * ((2 *n + 1) *(n^2 + n). The denominators are simply 2^n. The 2nd part of the sequence is straightforward.

Guest May 12, 2020