1.

Prove by pmi 1³+2³+3³+...n³=n³(n+1)³/4​

Answer»

Answer:

for n = 1

LHS = 1³ = 1

RHS = [ 1² (1 + 1)² ] / 4 = 4/4 = 1

which is TRUE

SUPPOSE that n = k is true

then

1³ + 2³ + 3³ + ... + k³ = [ k² (k + 1)² ] /4

now suppose that n = k + 1 is true

LHS

= 1³ + 2³ + 3³ + ... + k³ + (k + 1)³

=[ k² (k + 1)² ] /4 + (k + 1)³

= [ k² (k + 1)² + 4(k + 1)³ ] /4

= [ k² (k + 1)² + 4(k + 1)³ ] /4

= [ k⁴ + 2k³ + k² + 4k³ + 12k² + 12k + 4 ] /4

= [ k⁴ + 6k³ + 13k² + 12k + 4 ] /4

RHS

= [ (k + 1)² (k + 2)² ] /4

= [ (k² + 2k + 1)(k² + 4k + 4) ] /4

= [ k⁴ + 6k³ + 13k² + 12k + 4 ] /4

LHS = RHS

P(k) = P(k + 1)

so by mathematical INDUCTION, the statement is true


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