Question

4³+8³+12³+.............+up to n terms =16n²(n+1)² by

Answer 1

We have to prove that 4³ + 8³ + 12³ + ..... + upto n terms = 16n²(n + 1)² by mathematical induction.

Let P(n) = 4³ + 8³ + 12³ + ..... + upto n terms = 16n²(n + 1)²

for n = 1,

LHS = P(1) = 4³ = 64

RHS = 16(1)²(1 + 1)² = 16 × 4 = 64

LHS = RHS

so, P(1) is true.

for n = k,

P(k) = 4³ + 8³ + 12³ + .... + (4k)³ = 16k²(k + 1)²........(1)

for n = (k + 1)

P(k + 1) = 4³ + 8³ + 12³ + .... + {4(k + 1)}³ = 16(k + 1)²(k + 2)² .........(2)

from equation (1)

4³ + 8³ + 12³ + .... + (4k)³ = 16k²(k + 1)²

adding {4(k + 1)}³ equation (1) we get,

4³ + 8³ + 12³ + ....... + (4k)³ + {4(k + 1)}³ = 16k²(k + 1)² + {4(k + 1)}³

= 16k²(k + 1)² + 64(k + 1)³

= 16(k + 1)²[k² + 4k + 4]

= 16(k + 1)²(k + 2)²

Therefore 4³ + 8³ + 12³ + ....... + (4k)³ + {4(k + 1)}³ = 16(k + 1)²(k + 2)²

from equations (2) , P(K+1) is true . it is only when P(k) is true. from principle of mathematical induction, P(n) is true for all natural number.

Therefore 4³ + 8³ + 12³ + ..... upto n terms = 16n²(n + 1)² is true.

hence proved.