Sum disturbance method, find a compact form of the sum

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I tried to solve this example, but without success, could someone help me solve it because I got stuck on it and don't understand how to solve it.
Using the sum disturbance method, find a compact form of the following sum:

[tex]\textrm{(a)} \sum_{k=1}^n{(1+k2^{k-1})^2}[/tex]

Spoiler: show

The disturbance method is sum
[tex]{s_{n+1} = a_{1} + a_2 + \dots + a_n + a_{n+1}}[/tex]
expressed in two ways. The first is obvious:
[tex]s_{n+1} = s_n + a_{n+1}.[/tex]
The second is to present [tex]s_{n+1}[/tex] in form:
[tex]s_{n+1} = a_1 + f(s_n),[/tex]
where f is a function. Then we get the equality
[tex]s_n + a_{n+1} = a_1 + f(s_n).[/tex]
This equality can be treated as an equation with one unknown sn. When it is solved in relation to this unknown, we obtain a compact form of a sum.

[shyamjayakannan]

[tex]S_{n+1}=S_n+\left\{1+(n+1)2^n\right\}^2=4+\sum_{k=2}^{n+1}\left(1+k2^{k-1}\right)^2=4+\sum_{k=1}^{n}\left\{1+(k+1)2^k\right\}^2=4+\sum_{k=1}^{n}\left\{1+k2^{k-1}\times2+2^k\right\}^2[/tex]

[tex]=4+\sum_{k=1}^{n}\left\{2\left(1+k2^{k-1}\right)+2^k-1\right\}^2=4+\sum_{k=1}^n\left\{4\left(1+k2^{k-1}\right)^2+4\left(1+k2^{k-1}\right)\left(2^k-1\right)+\left(2^k-1\right)^2\right\}[/tex]

[tex]=4+4\sum_{k=1}^n\left(1+k2^{k-1}\right)^2+\sum_{k=1}^n\left\{4\left(1+k2^{k-1}\right)\left(2^k-1\right)+\left(2^k-1\right)^2\right\}=4+4S_n+\sum_{k=1}^n\left(2^{k+2}-4+k2^{2k+1}-k2^{k+1}+2^{2k}-2^{k+1}+1\right)[/tex]

[tex]=4+4S_n+\sum_{k=1}^n\left(2^{k+1}+2^{2k}\right)+\sum_{k=1}^n3+\sum_{k=1}^n\left(k2^{2k+1}-k2^{k+1}\right)=4+4S_n+4(2^n-1)+\frac{4}{3}(2^{2n}-1)+3n+2\sum_{k=1}^nk4^k-2\sum_{k=1}^nk2^k...(1)[/tex]

Let us now look at [tex]\sum_{k=1}^nk2^k[/tex]. Let this be = [tex]S_n'[/tex]. So, [tex]S_{n+1}'=S_n'+(n+1)2^{n+1}=2+\sum_{k=2}^{n+1}k2^k=2+\sum_{k=1}^n(k+1)2^{k+1}=2+2\sum_{k=1}^nk2^k+\sum_{k=1}^n2^{k+1}[/tex]

[tex]=2+2S_n'+4(2^n-1)\Rightarrow S_n'+(n+1)2^{n+1}=2S_n'+2^{n+2}-2\Rightarrow S_n'=\sum_{k=1}^nk2^k=2\left(n2^n-2^n+1\right)[/tex]

Similarly, [tex]\sum_{k=1}^nk4^k=\frac{4}{9}\left(3n4^n-4^n+1\right)[/tex]. Substituting these into (1), gives:

[tex]S_n+\left\{1+(n+1)2^n\right\}^2=4+4S_n+4(2^n-1)+\frac{4}{3}(2^{2n}-1)+3n+\frac{8}{9}\left(3n4^n-4^n+1\right)-4\left(n2^n-2^n+1\right)[/tex]

Finally, [tex]\boxed{S_n=\frac{n^24^n}{3}+n2^{n+1}-\frac{n2^{2n+1}}{9}+n+\frac{5\times4^n}{27}-2^{n+1}+\frac{49}{27}}[/tex]