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Pseudocode - painting a wall

My pseudocode from Module Two: Pseudocode – Painting a Wall MEASURE wall-length MEASURE wall width CALCULATE wall area DETERMINE amount of paint needed for wall-size PURCHASE desired paint and tools (brushes, rollers, pan, etc.) PREPARE work area and tools START painting wall END when painting is complete CLEAN work area and tools STORE tools and any leftover pa​‌‌‌‌‌‌‍‍‍‌‌‍‍‍‌‍‌‍‍​int

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Question: prove by induction 2^2 + 4^2 + 6^2 + ... + (2n)^2 = (2n)(2n+1)(2n+2)/6 ANSWER we will use induction on n base case : n=1 we have, 2^2 = 2*3*4/6 = 4 which is true inductive hypothesis let it be true for n = k i.e.,  2^2 + 4^2 + ... + (2k)^2 =   [(2k)(2k+1)(2k+2)]/6 inductive case let n = k+1 then we have 2^2 + 4^2 + .... + (2k)^2 + (2(k+1))^2 =   [(2k)(2k+1)(2k+2)]/6 + (2k+2)^2 =(2k+2)*[(2k)(2k+1)/6 + (2k+2)] =(2k+2)*[ (4k^2+2k)/6 + (12k + 12)/6 ] =(2k+2)*[ (4k^2+14k+12)/6 ] = =(2k+2)*[(2k)(2k+1)/6 + (2k+2)] =(2k+2)*[ (4k^2+2k)/6 + (12k + 12)/6 ] =(2k+2)*[ (4k^2+14k+12)/6 ] = (2k+2)*[ (4k^2 + 8k + 6k + 12)/6 ] = (2k+2)*[ (4k(k + 2) +6(k+2))/6 ] = (2k+2)*[ (4k+6)(k+2)/6 ] =  (2k+2)*[ 2 (2k+3)(k+2)/6  ] =   (2k+2)*[  (2k+3)*2*(k+2)/6  ] =   (2k+2)*[  (2k+3)(2k+4)/6  ] = [(2*(k+1))(2*(k+1)+1)(2*(k+1)+2)]/6 replacing k+1 by m, we get replacing k+1 by m, we get [(2*m)(2*m+1)(2*m+2)]/6 this completes our proof b...
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Question: Q1 a. Sketch the static characteristics of a diode. (6 marks) b. An AC voltage source V drives a transformer with centre tap second indin gs connected to two rectifiers is shown in the following diagram. The load has constant current Io. Assume that when the diode conducts the forward voltage drop is Ve d V Vmsnot. The turns ratio is N: 1:1. Sketch the current waveform through one of the diodes. (6 marks) c. Write down an equation for efficiency of a power converter in terms of output power and converter losses (2 marks) d. Given the following parameters, calculate the efficiency of the power converter. Va = 310 V ω-2π50 forward voltage drop of the diodes VF = 0.7 V Transformer turns ratio = 10: 1 : 1 load current = 10 A Assume that the transformer has no loss and converter losses come from the diodes only. Also assume the diode voltage drop does not change with current. State your assumption if there is any 6 marks) ANSWER
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