Question

Please refer to image! Indices question

Answer 1

k

Answer 2

Help with Q12! :)

Answer 3

\[P=(2+\sqrt{5})^{2n-1}\]

\[P=\frac{ (2+\sqrt{5})^{2n}(2-\sqrt{5}) }{ (2+\sqrt{5})(2-\sqrt{5}) }\]

\[P=\frac{ [(2+\sqrt{5})^{2}]^{n}(2-\sqrt{5}) }{ 4-5 }\]

Answer 4

can you try now the next step?

Answer 5

@BelleFlower

Answer 6

\[P=\frac{ (9+4\sqrt{5})^{n}(2+\sqrt{5}) }{ -1 }\]
P=\[P=(\sqrt{5}-2)(9+4\sqrt{5})^{n}\]

Answer 7

you can continue the next step

Answer 8

thats ok :D

Answer 9

check on this one
\[\sqrt{(4)^{2}}\sqrt{5}=\sqrt{16*5}=\sqrt{80}\]

Answer 10

OH OH okay thank you so much! I got it now!!

Answer 11

YW good luck now ... :D