Recurrence Relation

Question

How to solve this question ???  it ws getting lengthy . I tried to evaluate the nth term but I was stuck at some point. Please help 

Comments

assume particular solutn,  $a_{n}=(cn+d).2^{n}$

substitute it in a main eqtn, you will get,

$(cn+d).2^{n}+(c(n-1)+d).2^{n-1}=3n.2^{n}$

$\Rightarrow 3cn+3d-2=6n$

$\Rightarrow 3cn-6n+3d-2=0$

now above eqn should always hold true irrespective of value of 'n'.

$\Rightarrow 3c=6$ ,$\Rightarrow c=2$

putting value of c=2 in above, we get $\Rightarrow d=\frac{2}{3}$

so required particular solution is $a_{n}=(2n+\frac{2}{3}).2^{n}$

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Thank you so much @joshi_nitish

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Can you please provide the theory required to solve this question?