Question

find p

Answer 1

\[f(n)=15^n-8^{n-2}\]
and
\[f(n+1)-8f(n)=15^np\]

Answer 2

7 ?

Answer 3

\[f(n+1)-8f(n)=(f(n)+8^{n-2})p\]

\[f(n+1)-8^{n-2}p=f(n)(p+8)\]
lets try \[f(0)\]
\[f(0)=15^0-8^{0-2}=1-\frac{1}{64}=\frac{63}{64}\]
\[f(1)=15-8^{1-2}=15-\frac{1}{ 8}=\frac{114}{8}\]
\[f(1)-\frac{p}{64}=f(0)(p+8)\]
\[\frac{114}{8}-\frac{p}{64}=\frac{63}{64}(p+8)\]

Answer 4

thats ho far i went

Answer 5

\[114*8-p=63p+63*8\]
\[64p=8*114-63*8\]
\[64p=51*8\]
p=6.37

Answer 6

\[f(n)=15^n-8^{n-2}\]
\[f(n+1)=15^{n+1}-8^{n+1-2}\]

Answer 7

ohhhhh thats easier

Answer 8

\[\frac{15^{n+1}-8^{n-1}-8(15^{n}-8^{n-2})}{15^n}=p\]
\[\frac{15^{n+1}-8^{n-1}-8(15^{n})+8^{n-1})}{15^n}=p\]
\[\frac{15^{n+1}-8(15^{n})}{15^n}=p\]
\[15-8=p=7\]

Answer 9

alternative :
f(n)=15^n−8^(n−2)
if n = 0 then f(0) = 1 - 1/64 = 63/64
if n = 1 then f(1) = 15 - 1/8 = 119/8

f(n+1)−8f(n)=15np
put n=0 to get
f(1) - 8f(0) = 15^n p
119/8 - 8*63/64 = 15^0 p
119/8 - 63/8 = p
p = 56/8 = 7

Answer 10

* f(n+1)−8f(n)=15^n p

Answer 11

wow okay thanks guys for the time ...medals!!!

Answer 12

welcome :)

Answer 13

thank you