Question

t1=8, t2=32, tn=2tn-1 + 3tn-2 for integers >=3
use induction to prove that
tn=2 * (-1)^n + 10 * 3^(n-1)
@MIT 18.02 Multiva…

Answer 1

i've proved that n=3 is true
but i'm having trouble with adding in the n+1, i dont know what i'm suppose to add in

Answer 2

well they give you tn correct

Answer 3

set tn = to what they want you to prove

Answer 4

and then replace all n=k and move one summation up or do t(k)+t(k+1)=

Answer 5

i think in this case you just have to think of t as a constant

Answer 6

now is it like it is or is there like exponents and stuff in the first part?

Answer 7

no there are no exponents in teh first part. what shoudl t(k+1) be?

Answer 8

\[t(n+1)=(2t(n+1)-1)+(3t(n+1)-2)\]

Answer 9

but how would i prove it then..... it becomes:
2*(-1)^n+10*3^(n-1)+2tn+3t(n-1)

Answer 10

i have no idea how to go from there

Answer 11

i wonder if you set tn to = a single variable

Answer 12

something like tn=k=2k-1+3k-2

Answer 13

try working backwards... try making your right side = tn

Answer 14

\[t_{n+1}=2t_{n} + 3t_{n-1}\]

\[t_{n+1}=2(2 (-1)^{n} + 10 \cdot 3^{n-1}) + 3\cdot(2 (-1)^{n-1} + 10 \cdot 3^{n-2})\]

\[=4 (-1)^{n} + 20 \cdot 3^{n-1} + 6 (-1)^{n-1} + 30 \cdot 3^{n-2}\]
\[=4 (-1)^{n} + 6 (-1)^{n-1} + 20 \cdot 3^{n-1} + 30 \cdot 3^{n-2}\]
\[=-4 (-1)^{n-1} + 6 (-1)^{n-1} + 20 \cdot 3^{n-1} + 10 \cdot 3^{n-1}\]
\[=2 (-1)^{n-1} + 30 \cdot 3^{n-1} \]
\[=2 (-1)^{n-1}\cdot(-1)^2 + 10\cdot3 \cdot 3^{n-1} \]
\[=2 (-1)^{n+1} + 10\cdot 3^{n} \]