## satellite73 Group Title find the first 4 terms $a_1=\frac{3}{2}; a_{n+1}=\frac{n^2+1}{n(a_n)}$ one year ago one year ago

1. satellite73 Group Title

i got my answer but apparently i was wrong this from @lopez_hatesmath

2. satellite73 Group Title

maybe some fresh eyes would help

3. satellite73 Group Title

nvm sorry to bother you

4. lopez_hatesmath Group Title

:(

5. satellite73 Group Title

i got it

6. lopez_hatesmath Group Title

swagg

7. satellite73 Group Title

it was this $a_1=\frac{3}{2}; a_{n+1}=\frac{n^2+1}{n}\times a_n$

8. satellite73 Group Title

that is, the $$a_n$$ was in the NUMERATOR

9. satellite73 Group Title

so now it is not so bad replace $$n=1$$ on the right hand side to get $$a_2$$

10. satellite73 Group Title

you get $a_2=\frac{1^2+1}{1}\times \frac{3}{2}$ $a_2=2\times\frac{3}{2}$ $a_2=3$

11. satellite73 Group Title

now replace $$n$$ by 2 on the right hand side to get $a_3=\frac{2^2+1}{2}\times 3$ $a_3=\frac{5}{2}\times 3$ $a_3=\frac{15}{2}$

12. satellite73 Group Title

how are we doing so far?

13. lopez_hatesmath Group Title

good :)

14. satellite73 Group Title

one more? $a_4=\frac{3^2+1}{3}\times \frac{15}{2}$ $a_4=\frac{10}{3}\times \frac{15}{2}$ $a_4=25$

15. lopez_hatesmath Group Title

okayy i got it!

16. satellite73 Group Title

not so bad that is the idea, i thought the term was in the denominator which is why i was screwing it up

17. lopez_hatesmath Group Title

gotcha thanks for the help man.

18. satellite73 Group Title

yw good luck with the next one, but it works the same so you should be good to go