document.write( "Question 26478: Use the principle of mathematical induction to prove the following identity.
\n" ); document.write( "If n is greator and equal to 1 then, \r
\n" ); document.write( "\n" ); document.write( "1*2+.......+n(n+1) = (n(n+1)(n+2))/3 \n" ); document.write( "
Algebra.Com's Answer #14420 by venugopalramana(3286)\"\" \"About 
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1*2+.......+n(n+1) = (n(n+1)(n+2))/3
\n" ); document.write( "IN INDUCTION ,WE FIRST TEST WHETHER IT IS CORRECT FOR N=1
\n" ); document.write( "LHS=1*2=2
\n" ); document.write( "RHS=1*2*3/3=2...LHS=RHS...SO IT IS TRUE FOR N=1
\n" ); document.write( "NOW WE ASSUME IT IS TRUE FOR N=K ......SOME VALUE...SO
\n" ); document.write( "1*2+.......+K(K+1) = (K(K+1)(K+2))/3........................I
\n" ); document.write( "WE NOW GO ON TO PROVE THAT IT IS TRUE FOR N=K+1...THAT IS...TPT
\n" ); document.write( "1*2+.......+K(K+1)+(K+1)(K+2) = ((K+1)(K+2)(K+3))/3..........II\r
\n" ); document.write( "\n" ); document.write( "LHS={1*2+.......+K(K+1)}+(K+1)(K+2)...USING EQN.I
\n" ); document.write( "=(K(K+1)(K+2))/3}+(K+1)(K+2)
\n" ); document.write( "=(K+1)(K+2){(K/3)+1}
\n" ); document.write( "=(K+1)(K+2){(K+3)/3}
\n" ); document.write( "=((K+1)(K+2)(K+3))/3=RHS OF EQN.II.
\n" ); document.write( "SO WE GOT IT NOW THAT ,IF THE GIVEN RELATION IS TRUE FOR N=K,THEN IT IS TRUE FOR N=K+1
\n" ); document.write( "BUT WE PROVED FIRST THAT IT IS TRUE FOR N=1
\n" ); document.write( "SO IT IS TRUE FOR N=2
\n" ); document.write( "SO IT IS TRUE FOR N=3
\n" ); document.write( "SO IT IS TRUE FOR N=4
\n" ); document.write( "........ETC..........
\n" ); document.write( "SO IT IS TRUE FOR ALL INTEGRAL VALUES OF N\r
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