Asked by A-A
1(2)+2(3)+3(4)+...+n(n+1)= n(n+1)(n+2)/2
using the mathematical induction.
please help..
using the mathematical induction.
please help..
Answers
Answered by
MathMate
1(2)+2(3)+3(4)+...+n(n+1)= n(n+1)(n+2)/<b>3</b>
(Note: The denominator is 3, not 2)
Basis: n=1
1(2)=2
1(2)(3)/3=2
n=1 is valid.
Assume:
1(2)+2(3)+3(4)+...+k(k+1)= k(k+1)(k+2)/3
is valid for k=n
then try to prove that the relation is valid for k=n+1.
1(2)+2(3)+3(4)+...+n(n+1) + (n+1)(n+2)
= n(n+1)(n+2)/3 + (n+1)(n+2)
= ( n(n+1)(n+2)+3(n+1)(n+2) )/3
= ( n^3+6*n^2+11*n+6 )/3
= (n+1)(n+2)(n+3)/3
= (n+1)*(n+1 +1)*(n+1 +2)/3
which means that the expression is valid also for n+1. QED
(Note: The denominator is 3, not 2)
Basis: n=1
1(2)=2
1(2)(3)/3=2
n=1 is valid.
Assume:
1(2)+2(3)+3(4)+...+k(k+1)= k(k+1)(k+2)/3
is valid for k=n
then try to prove that the relation is valid for k=n+1.
1(2)+2(3)+3(4)+...+n(n+1) + (n+1)(n+2)
= n(n+1)(n+2)/3 + (n+1)(n+2)
= ( n(n+1)(n+2)+3(n+1)(n+2) )/3
= ( n^3+6*n^2+11*n+6 )/3
= (n+1)(n+2)(n+3)/3
= (n+1)*(n+1 +1)*(n+1 +2)/3
which means that the expression is valid also for n+1. QED
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