Question

plse help me to solve this

Answer 1

1) Does it start? Try P(0).
2) Does one imply the next? P(n) + Term(n+1) = P(n+1)

Go!

Answer 2

i have solve it
but not getting answer when i put P(n+1)

Answer 3

Not good with PMI bro. Sry. :/

Answer 4

okay

Answer 5

\[P(n+1) = 3 . 2^2 + 3^2 . 2^3 +....+ 3^n . 2^{n+1} + 3^{n+1} . 2^{n+2}\]
We have assumed P(n) is true, which gives the below step\[P(n+1) = \frac{ 12(6^n - 1) }{ 5 } + 3^{n+1} . 2^{n+2}\]
\[P(n+1) = \frac{12(6^n - 1) + 5 . 3^{n+1} . 2^{n+1} }{ 5 }\] \[P(n+1) = \frac{ 12(6^n - 1) + 5 . 3^n.3.2^n.2^2 }{ 5 }\] \[P(n+1) =\frac{ 12(6^n - 1 + 5.3^n.2^n) }{ 5 }\]\[P(n+1) = \frac{ 12(6^n - 1 + 5.6^n) }{ 5 }\]\[P(n+1) = \frac{ 12(6.6^n - 1) }{ 5 }\]\[P(n+1) = \frac{ 12(6^{n+1} - 1) }{ 5 }\]

Answer 6

Oops.. typo in the 3rd equation. Its not \[2^{n+1}\] its \[2^{n+2}\]