+0

# HELP MATH HW

0
290
6
+14

1) Given that $$a_0 = 1$$ and $$a_1 = 5$$, and the general relation $$a_n^2 - a_{n - 1} a_{n + 1} = (-1)^n$$

for $$n \ge 1,$$ find $$a_{10}.$$

2)Evaluate the sum. $$\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots.$$

3)$$A sequence \{a_n\} satisfies a_1 = 1 and $a_n = \frac{a_{n - 1}}{1 + a_{n - 1}}$for all n \ge 2. Find a_{10}.$$

4)The sum of the first n terms of a certain sequence is n(n+1)(n+2) Find the tenth term of the sequence.

Mar 9, 2019

#1
+8336
+1

1.

(Wow, this is some really advanced homework)

The generated sequence is 1,5,26,135,701,..., which is OEIS A052918 (Took me a while to find it). And a10 is 2,646,275.

Notes: The general relation can be rewritten as an+1 = 5an + an-1.

https://oeis.org/A052918

^ Here's a link to the sequence.

Mar 9, 2019
#4
+14
+1

Thanks for the two answers but for the link on the first one I am confused on how to read it.

SUS101  Mar 9, 2019
#6
+8336
+1

You read the line that says "A052918              a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1)."

That's the index of the sequence and the recursive formula for it.

Then the list below are the members of the sequence.

MaxWong  Mar 9, 2019
#2
+8336
+2

2.

$$\dfrac{6}{3^2-1} + \dfrac{6}{5^2-1} +...\\ = \displaystyle 6\sum^{\infty}_{n=0} \dfrac{1}{(2n+3)^2 - 1}\\ = \displaystyle \dfrac{3}{2}\sum^{\infty}_{n=0} \dfrac{1}{(n+1)(n+2)}\\ = \displaystyle \dfrac{3}{2}\sum^{\infty}_{n=0} \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)\\ = \displaystyle \dfrac{3}{2}\left(\sum^{\infty}_{n=0} \dfrac{1}{n+1} - \sum^{\infty}_{n=0}\dfrac{1}{n+2}\right)\\ = \displaystyle \dfrac{3}{2}\left(\sum^{\infty}_{n=0} \dfrac{1}{n+1} - \sum^{\infty}_{n=1}\dfrac{1}{n+1}\right)\\ = \dfrac{3}{2}\left(\dfrac{1}{0+1}\right)\\ =\dfrac{3}{2}$$

Mar 9, 2019
#3
+8336
+2

3.

$$a_1 = 1\\ a_2 = \dfrac{a_1}{1+a_1} = \dfrac{1}{2}\\ a_3 = \dfrac{a_2}{1+a_2} = \dfrac{1}{3}\\ a_4 = \dfrac{a_3}{1+a_3} = \dfrac{1}{4}\\ \vdots$$

Obviously, $$a_n = \dfrac{1}{n}\quad \forall n \geq 1$$.

$$\therefore a_{10} = \dfrac{1}{10}$$

Mar 9, 2019
#5
+8336
+2

4.

$$\displaystyle \sum^{n}_{k=1} a_k = n(n+1)(n+2)$$.

$$a_{10} = \displaystyle \sum^{10}_{k=1} a_k - \sum^9_{k=1} a_k = 10(11)(12) - 9(10)(11) = 330$$

.
Mar 9, 2019
edited by MaxWong  Mar 9, 2019