3 years ago

Please Help! It’s Due TODAY!!

Mathematics

1 answer:

dalvyx [7]3 years ago

3 0

Answer:

4096

256

1024

Step-by-step explanation:

i literally just looked these up sorry

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Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D

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By iteratively substituting, we have

$a_n = a_{n-1} + n$

$a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)$

$a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)$

and the pattern continues down to the first term $a_1=0$ ,

$a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))$

$\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)$

$\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k$

Recall the formulas

$\displaystyle \sum_{n=1}^N 1 = N$

$\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2$

It follows that

$a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2$

$\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1$

$\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}$

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2 years ago