Question

prove that for every natural number n

Answer 1

\[25| 2^{n+2}3^n+5n-4\]

Answer 2

??

Answer 3

\[a|b \] means b is divisible by a

Answer 4

but what to prove ?

Answer 5

look at these
http://openstudy.com/users/jonask#/updates/50574352e4b02986d3712444

Answer 6

http://openstudy.com/users/jonask#/updates/50560ad9e4b0a91cdf44aa8f

Answer 7

i believe u can apply induction

Answer 8

\[n=k\]
let \[4*6^k+5k-4\]
\[k=1\]
\[4*6+5-4=25\]
true

Answer 9

\[n=k+1\]
\[LHS=24*6^k+5k+1\]

Answer 10

note that\[P(k+1)=2^{k+3}3^{k+1}+5(k+1)-4\]\[=6\times (2^{k+1}3^k+5k-4)-6(5k-4)+5(k+1)-4\]

Answer 11

how does it help us

Answer 12

oh wait its the same thing as we substitute n=k

Answer 13

isnt it supposed t be 12 instead of 6

Answer 14

\[P(k+1)=6\times P(k)-25k+25\]

Answer 15

oh thanks,i see

Answer 16

sorry :/