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Answer:
1 3 + 2 3 + 3 3 + ... + n 3 = n 2 (n + 1) 2 / 4
STEP 1: We first show that p (1) is true.
Left Side = 1 3 = 1
Right Side = 1 2 (1 + 1) 2 / 4 = 1
hence p (1) is true.
STEP 2: We now assume that p (k) is true
1 3 + 2 3 + 3 3 + ... + k 3 = k 2 (k + 1) 2 / 4
add (k + 1) 3 to both sides
1 3 + 2 3 + 3 3 + ... + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3
factor (k + 1) 2 on the right side
= (k + 1) 2 [ k 2 / 4 + (k + 1) ]
set to common denominator and group
= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4
= (k + 1) 2 [ (k + 2) 2 ] / 4
We have started from the statement P(k) and have shown that
1 3 + 2 3 + 3 3 + ... + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4
3 + 2 3 + 3 3 + ... + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4 Which is the statement P(k + 1).
Answer:
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