Question 333969
in this series,  {{{a[i]=a[i-1]  +4^(i-2)}}} where {{{a[0]=3/4}}} and i>0
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for i=1,  {{{a[1]=a[0]+4^-1=(3/4)+(1/4)=1}}}
for i=2,  {{{a[2]=a[1]+4^0=1+1=2}}}
for i=3,  {{{a[3]=a[2]+4^1=2+4=6}}}
for i=4,  {{{a[4]=a[3]+4^2=6+16=22}}}
for i=5,  {{{a[5]=a[4]+4^3=22+64=86}}}
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for i=6,  {{{a[6]=a[5]+4^4=86+256=342}}}
for i=7,  {{{a[7]=a[6]+4^5=342+1024=1366}}}
for i=8,  {{{a[8]=a[7]+4^6=1366+4096=5462}}}