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 caselet n = k+1then we have2^2 + 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 getreplacing k+1 by m, we get[(2*m)(2*m+1)(2*m+2)]/6this completes our proof by induction
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 caselet n = k+1then we have2^2 + 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 getreplacing k+1 by m, we get[(2*m)(2*m+1)(2*m+2)]/6this completes our proof by induction
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