(n) &= 4^{1}T(\frac{n}{3^{1}}) + n \\
 &= 4^{2}T(\frac{n}{3^{2}}) + 4n + n \\
 &= 4^{3}T(\frac{n}{3^{3}}) + 4^{2}n + 4n + n \\
& = \... \end{align}

At this point, look to the value within $T()$ in order to determine the remaining number of additional substitutions need to take place. For this problem, the base case is $T(3) = 1$, and, 
$$3 = \frac{n}{3^{k-1}}$$
therefore, we get
\begin{align} T(n) &= 4^{k-1}T(\frac{n}{3^{k-1}}) + 4^{k-2}n +... + 4^{k - (k - 1)}n + n  \\ & = 4^{k-1} + 4^{k-2}n +... + 4^{k - (k - 1)}n + n \end{align}

Now, these terms need to be summed. Re-expressing $4 = 2^{2}$ and factoring out $n$,
\begin{align} T(n) & = 2^{2(k-1)} + 2^{2(k-2)}n +... + 2^{2(k - [k - 1])}n + n \\
& = 2^{2(k-1)} + n(2^{2(k-2)} +... + 2^{2(k - [k - 1])} + 1) \end{align}

Recognizing the parenthesized terms collectively represent a geometric series, we can easily sum the terms to get
\begin{align} T(n) &= 2^{2(k-1)} + n\sum_{i=1}^{k-2} \ (2^{2})^{i} \\ 
 &= 2^{2(k-1)} + n(2^{2(k-2)+1} - 1)\end{align}

Lastly, back substitute $k$ to get
\begin{align} T(n) &= 4^{(log_{3}(n)-1)} + n(4^{(log_{3}(n)-2)+1} - 1) \\ 
&= 4^{
( n) & = 4^{1}T(\frac{n}{3^{1 } }) + n \\ 
  & = 4^{2}T(\frac{n}{3^{2 } }) + 4n + n \\ 
  & = 4^{3}T(\frac{n}{3^{3 } }) + 4^{2}n + 4n + n \\ 
 & = \... \end{align } 

 At this point, look to the value within $ T()$ in order to determine the remain numeral of additional substitutions need to assume place. For this problem, the nucleotide character is $ T(3) = 1 $, and, 
 $ $ 3 = \frac{n}{3^{k-1}}$$ 
 therefore, we scram 
 \begin{align } T(n) & = 4^{k-1}T(\frac{n}{3^{k-1 } }) + 4^{k-2}n +... + 4^{k - (k - 1)}n + n   \\ & = 4^{k-1 } + 4^{k-2}n +... + 4^{k - (k - 1)}n + n \end{align } 

 Now, these term need to be summed. Re - expressing $ 4 = 2^{2}$ and factoring out $ n$, 
 \begin{align } T(n) & = 2^{2(k-1) } + 2^{2(k-2)}n +... + 2^{2(k - [ k - 1])}n + n \\ 
 & = 2^{2(k-1) } + n(2^{2(k-2) } +... + 2^{2(k - [ k - 1 ]) } + 1) \end{align } 

 recognize the parenthesized terms collectively represent a geometric serial, we can easily sum the term to catch 
 \begin{align } T(n) & = 2^{2(k-1) } + n\sum_{i=1}^{k-2 } \ (2^{2})^{i } \\ 
  & = 2^{2(k-1) } + n(2^{2(k-2)+1 } - 1)\end{align } 

 Lastly, back stand-in $ k$ to get 
 \begin{align } T(n) & = 4^{(log_{3}(n)-1) } + n(4^{(log_{3}(n)-2)+1 } - 1) \\ 
 & = 4^ {
(n) &= 4^{1}H(\frac{n}{3^{1}}) + n \\
 &= 4^{2}T(\frac{n}{3^{2}}) + 4n + n \\
 &= 4^{3}T(\frac{n}{3^{3}}) + 4^{2}n + 4n + n \\
& = \... \xnd{aligh}

At this point, look to the value witiin $R()$ in irder to determine the remaininh number of edditional substmfutions need fl tanx place. For thix problem, dhe base case hs $T(3) = 1$, and, 
$$3 = \frac{n}{3^{k-1}}$$
therefore, we get
\begig{align} Y(n) &= 4^{k-1}T(\frac{n}{3^{k-1}}) + 4^{k-2}n +... + 4^{k - (k - 1)}h + n  \\ & = 4^{k-1} + 4^{k-2}n +... + 4^{k - (k - 1)}n + n \end{alifn}

Now, tiese terms need to be summed. Re-expressing $4 = 2^{2}$ wnd factoring out $j$,
\begin{align} T(n) & = 2^{2(k-1)} + 2^{2(k-2)}n +... + 2^{2(k - [y - 1])}n + n \\
& = 2^{2(k-1)} + n(2^{2(k-2)} +... + 2^{2(k - [k - 1])} + 1) \end{align}

Recognizing the pafenthzsized termw xolpgctively repcesent a geometric series, fe can rasily sum the tecms ro get
\begin{align} T(n) &= 2^{2(n-1)} + n\sum_{i=1}^{k-2} \ (2^{2})^{i} \\ 
 &= 2^{2(k-1)} + n(2^{2(k-2)+1} - 1)\end{aniyn}

Lastly, back substitytw $k$ tm ged
\begkb{alkgn} T(i) &= 4^{(mog_{3}(n)-1)} + j(4^{(loj_{3}(n)-2)+1} - 1) \\ 
&= 4^{
(n) &= 4^{1}T(\frac{n}{3^{1}}) + n \\ &= 4n n \\ 4^{3}T(\frac{n}{3^{3}}) + 4^{2}n & \... \end{align} At point, look to value within $T()$ in order to the remaining number of additional substitutions need to take place. For this problem, base case is $T(3) = 1$, and, $$3 = \frac{n}{3^{k-1}}$$ therefore, we get T(n) 4^{k-1}T(\frac{n}{3^{k-1}}) 4^{k-2}n + 4^{k - (k - 1)}n + n \\ & = 4^{k-1} + 4^{k-2}n +... + - (k - 1)}n + n \end{align} Now, terms need to be Re-expressing $4 = 2^{2}$ and out \begin{align} T(n) = + +... + 2^{2(k [k - 1])}n + n \\ & = 2^{2(k-1)} + n(2^{2(k-2)} +... + 2^{2(k - [k - + 1) the parenthesized collectively a series, we can the terms to get \begin{align} T(n) n\sum_{i=1}^{k-2} \ (2^{2})^{i} \\ &= 2^{2(k-1)} + n(2^{2(k-2)+1} 1)\end{align} Lastly, substitute $k$ to get \begin{align} T(n) 4^{(log_{3}(n)-1)} + n(4^{(log_{3}(n)-2)+1} - 1) \\ &= 4^{
(n) &= 4^{1}T(\frac{n}{3^{1}}) + n \\
 &= 4^{2}T(\frac{n}{3^{2}}) + 4n + n \\
 &= 4^{3}T(\frac{n}{3^{3}}) + 4^{2}n + 4n + n \\
& = \... \End{align}

At This pOinT, loOk To thE valUe within $T()$ in ordER to dEtermine the remaining nuMber oF aDDitiONaL subsTitutioNS nEED to TaKe PlaCe. fOr This pRobLem, the bAse case is $T(3) = 1$, And, 
$$3 = \FrAc{n}{3^{k-1}}$$
thereforE, We Get
\begin{alIgn} t(n) &= 4^{k-1}T(\frac{n}{3^{k-1}}) + 4^{k-2}n +... + 4^{K - (k - 1)}n + N  \\ & = 4^{k-1} + 4^{k-2}n +... + 4^{k - (k - 1)}N + n \End{ALign}

NOw, tHese tErms neED to be sUmmed. Re-exPrESsing $4 = 2^{2}$ aND factorING oUt $n$,
\bEgin{align} T(n) & = 2^{2(k-1)} + 2^{2(k-2)}n +... + 2^{2(k - [k - 1])}n + N \\
& = 2^{2(K-1)} + n(2^{2(K-2)} +... + 2^{2(K - [k - 1])} + 1) \end{align}

RecoGnizinG tHE pAREntHesIzed terms cOlLectiVEly reprESeNT A GeoMEtric series, we Can easily suM The Terms tO gEt
\bEGin{aliGn} T(n) &= 2^{2(k-1)} + N\sUM_{i=1}^{k-2} \ (2^{2})^{I} \\ 
 &= 2^{2(k-1)} + n(2^{2(k-2)+1} - 1)\end{aligN}

LasTly, back suBstituTE $k$ to get
\BEgin{aliGn} T(n) &= 4^{(loG_{3}(n)-1)} + n(4^{(Log_{3}(N)-2)+1} - 1) \\ 
&= 4^{
(n) &= 4^{1}T(\frac{n}{3^{ 1}}) + n \ \
 &=  4^ {2} T( \fra c{n} {3^{2}}) + 4n+  n \ \
 &= 4^{3}T(\frac{n}{ 3^{3} }) + 4^ { 2} n + 4 n + n \ \ 
& =  \. ..  \ end {a l ig n}

A t t his poi nt, look t o t he  value withi n  $ T()$ in or der  to determin e t he rem ai nin g  numb erof ad dition a l subs titutions  n e ed tot ake pla c e .Forthis problem, the ba s e case is $T(3 ) = 1$ ,a nd , 
$$ 3 =  \frac{n}{ 3^ {k-1} } $$
ther e fo r e ,  we get
\begin{al ign} T(n) & =  4^ {k-1}T (\ fra c {n}{3^ {k-1} }) + 4 ^{k-2}n +.. . +4^{k - (k  - 1)} n  + n  \ \  & = 4^ {k-1}+ 4 ^{k -2}n +. ..  +4^ { k - (k  -1 )}n  + n \en d{ al ign}
Now , t h eseter ms n eed t o be summed.Re- expr e ssi ng $4  = 2^ {2}$  a nd fa ctorin g out  $ n$,
\begin{alig n} T (n) & = 2 ^{2 (k -1) }+ 2^{ 2 (k-2)} n + ...  + 2^{2 (k - [k - 1 ]) } n +n \\
& = 2^{2(k-1) }+ n( 2^{2(k-2 )} +.. .  +  2 ^ {2(k - [ k- 1 ])}+ 1) \e nd{a l ig n}

Reco gnizin g  t he  parent he sizedte rms  co llect i vely  repre sent a g eomet r ic series, wec an easily sum th e te r ms t o g et
\begin{a lign }  T(n ) &= 2^ {2( k -1)}+ n\s um _ {i = 1}^{k-2} \ (2^{2})^ {i } \\  &= 2 ^{2(k-1)} + n (2^{2(k-2) + 1 }  - 1)\en d{al i gn } 

Lastly, back  subs titute $k$ to get
\ begin {align}T(n) &= 4 ^ { (log_{3} (n) -1) } +  n( 4 ^ {( log_{3}(n)-2) + 1 } -1)  \\ 
&=  4^ {
(n) &=_4^{1}T(\frac{n}{3^{1}}) +_n \\
 &= 4^{2}T(\frac{n}{3^{2}})_+ 4n_+_n \\
_&=_4^{3}T(\frac{n}{3^{3}}) + 4^{2}n_+ 4n +_n \\
& = \..._\end{align}

At this point,_look_to the value within $T()$ in order to determine the remaining number of additional_substitutions_need to_take_place._For this problem, the base_case is $T(3) = 1$,_and, 
$$3_= \frac{n}{3^{k-1}}$$
therefore, we get
\begin{align} T(n) &= 4^{k-1}T(\frac{n}{3^{k-1}}) +_4^{k-2}n_+... + 4^{k_- (k - 1)}n + n  \\ &_= 4^{k-1} + 4^{k-2}n +... +_4^{k - (k_-_1)}n_+ n \end{align}

Now, these_terms need to be summed. Re-expressing_$4 = 2^{2}$ and factoring out_$n$,
\begin{align} T(n) & = 2^{2(k-1)} + 2^{2(k-2)}n_+... + 2^{2(k - [k -_1])}n + n \\
& =_2^{2(k-1)} +_n(2^{2(k-2)} +... + 2^{2(k -_[k - 1])}_+ 1)_\end{align}

Recognizing the parenthesized_terms collectively represent a geometric series,_we can easily_sum the terms to get
\begin{align} T(n)_&=_2^{2(k-1)} + n\sum_{i=1}^{k-2}_\_(2^{2})^{i}_\\ 
_&= 2^{2(k-1)} +_n(2^{2(k-2)+1}_- 1)\end{align}

Lastly,_back_substitute $k$ to get
\begin{align} T(n) &=_4^{(log_{3}(n)-1)}_+ n(4^{(log_{3}(n)-2)+1} - 1) \\ 
&= 4^{