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Need answers for your academic questions? Ask them in the comment section below. Just type in the url link of your question in the comment section below. And get your answer ready in two formats. Hand written Screenshot We will post the answer as soon as we get them (questions) Thank you..
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...