document.write( "Question 773492: 1.3+3.5+5.7+....+(2n-1)(2n+1)=n(4n^2+6n-1)/3\r
\n" ); document.write( "\n" ); document.write( "prove by mathematical induction that
\n" ); document.write( "above statement holds true for every
\n" ); document.write( "integer n belongs to N.\r
\n" ); document.write( "\n" ); document.write( "HINT:to prove that (k+1)(4(k+1)^2+6(k+1)-1)/3 \n" ); document.write( "
Algebra.Com's Answer #471595 by pakhi(24)\"\" \"About 
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When n=1 we have the end term of the series as (2*1 -1)(2*1 +1) = 1*3 = 3
\n" ); document.write( "Putting n=1 in the r.h.s of the given equation we have
\n" ); document.write( "1(4*1^2 + 6*1 - 1)/3 = 1(4 + 6 -1)/3 = 3
\n" ); document.write( "Therefore the equation is valid for n=1
\n" ); document.write( "Let the expression be valid for any value n=k where 'k' belongs to N.
\n" ); document.write( "So 1.3 + 3.5 +.....+(2k-1)(2k+1)=k(4k^2+6k-1)/3 holds true. ---------eqn(1)
\n" ); document.write( "Now we have to prove that the equation is valid for n=k+1.
\n" ); document.write( "i.e. 1.3+3.5+....+{2(k+1)-1}{2(k+1)+1} = (k+1){4(k+1)^2+6(k+1)-1)/3----eqn(2)
\n" ); document.write( "Now l.h.s of equation 2 can be written as
\n" ); document.write( "1.3 + 3.5 +....+(2k-1)(2k+1) + {2(k+1)-1}{2(k+1)+1}-----------expression(1)
\n" ); document.write( "Putting the value of the r.h.s of equation 1 in expression 1 we have\r
\n" ); document.write( "\n" ); document.write( " k(4k^2+6k-1)/3 + (2k+2-1)(2k+2+1)
\n" ); document.write( "or (4k^3+6k^2-k)/3 + 3(2k+1)(2k+3)/3
\n" ); document.write( "or (4k^3+6k^2-k)/3 + 3(4k^2+6k+2k+3)/3
\n" ); document.write( "or (4k^3+6k^2-k)/3+ (12k^2+24k+9)/3
\n" ); document.write( "or (4k^3+6k^2-k+12k^2+24k+9)/3
\n" ); document.write( "or (4k^3+18k^2+23k+9)/3
\n" ); document.write( "or (4k^3+4k^2+14k^2+14k+9k+9)/3
\n" ); document.write( "or {4k^2(k+1)+14k(k+1)+9(k+1)}/3
\n" ); document.write( "or (k+1)(4k^2+14k+9)/3
\n" ); document.write( "or (k+1)(4k^2+8k+6k+4+6-1)/3
\n" ); document.write( "or (k+1){4k^2+8k+4+6k+6-1}/3
\n" ); document.write( "or (k+1){4(k^2+2k+1)+6(k+1)-1)/3
\n" ); document.write( "or (k+1){4(k+1)^2+6(k+1)-1}/3-----expression(2)
\n" ); document.write( "But expression 2 is nothing but the r.h.s. of the equation 2.
\n" ); document.write( "Therefore by mathematical induction we have proved that the said equation holds true for every value of 'n' where 'n' belongs to N.\r
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