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The following formula is used to calculate n: n = x * x/(1 - x) . Value x = 0 is used to stop algorithm. Calculation is repeated using values of x until value x = 0 is input. There is also a need to check for error conditions. Values of n and x must be output. Write an algorithm to display this repeated calculation using pseudocode.
NOTE: It's much easier in this illustration to input x first and then loop round doing calculation until eventually x = 0. Due to this, it would be essential to input x twice (which implies inside the loop and outside the loop). If input x occurred only once it would result in a more complicated algorithm. (Also note in algorithm that <> is used to represent ≠).
A while loop is used here, however a repeatloop would work just as well.
input x
while x <> 0 do
if x = 1 then print "error"
else n = (x * x)/(1 - x)
print n, x
endif
endwhile
for i=1 to n if a[i}>7 for j=2 to n a[j]=a{j}+j for n=2 to n a[k]=a[j]+i else if a[1]>4 && a[1] for 2 to a[1] a[j]= a{j]+5 else for 2to n a[j]=a[j]+i ..
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