Find the first six terms of the recursive sequence.

{eq}\displaystyle a_1 = 1,\ a_{n + 1} = 4 a_n - 2 {/eq}.

Given {eq}\displaystyle a_1=1 , a_{n+1}=4 a_n-2 {/eq}

Substituting {eq}\displaystyle n=1 {/eq}, we get

{eq}\displaystyle a_{1+1} =4a_1-2 \\ \Rightarrow a_2 = 4 (1)-2 \\ \Rightarrow a_2 =4-2 \\ \Rightarrow a_2 =2 {/eq}

Therefore, {eq}\displaystyle a_2=2 {/eq}

Substituting {eq}\displaystyle n=2 {/eq}, we get

{eq}\displaystyle a_{2+1} =4a_2-2 \\ \Rightarrow a_3 = 4 (2)-2 \\ \Rightarrow a_3 =8-2 \\ \Rightarrow a_3 =6 {/eq}

Therefore, {eq}\displaystyle a_3=6 {/eq}

Substituting {eq}\displaystyle n=3 {/eq}, we get

{eq}\displaystyle a_{3+1} =4a_3-2 \\ \Rightarrow a_4 = 4 (6)-2 \\ \Rightarrow a_4 =24-2 \\ \Rightarrow a_4 =22 {/eq}

Therefore, {eq}\displaystyle a_4=22 {/eq}

Substituting {eq}\displaystyle n=4 {/eq}, we get

{eq}\displaystyle a_{4+1} =4a_4-2 \\ \Rightarrow a_5 = 4 (22)-2 \\ \Rightarrow a_5 =88-2 \\ \Rightarrow a_5 =86 {/eq}

Therefore, {eq}\displaystyle a_5=86 {/eq}

Substituting {eq}\displaystyle n=5 {/eq}, we get

{eq}\displaystyle a_{5+1} =4a_5-2 \\ \Rightarrow a_6 = 4 (86)-2 \\ \Rightarrow a_6 =344-2 \\ \Rightarrow a_6 =342 {/eq}

Therefore, {eq}\displaystyle a_6=342 {/eq}

Therefore, the first six terms of the recursive sequence are {eq}\displaystyle \mathbf{a_1=1,a_2=2,a_3=6,a_4=22,a_5=86,a_6=342} {/eq}