mathematical induction 1+2+3+....+n=1/2(n+1) prove that (answer assignment)
RaunakRaj:
It should be n/2
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Given that, 1+2+3+........+n=n(n+1)/2......(1)
checking that p(1) is true.
then assume, p(k) is true:
1+2+3+....+k=k(k+1)/2
then check p(k+1) is true:
take LHS of the equ:
1+2+3+......+k+(k+1)=(k(k+1)/2)+(k+1)
{........................ .......=>[(1+2+3+...+k)+(k+1)]}
=k(k+1)/2+(k+1)
=(k^2+k+2k+2)/2
=(k^2+3k+2)/2 [:.k^2+3k+2=(k+1)(k+2)]
=(k+1)(k+2)/2//
so, p(k+1) is true.
and so, by the mathematical induction, we know that p(n) is true for every n£N.
checking that p(1) is true.
then assume, p(k) is true:
1+2+3+....+k=k(k+1)/2
then check p(k+1) is true:
take LHS of the equ:
1+2+3+......+k+(k+1)=(k(k+1)/2)+(k+1)
{........................ .......=>[(1+2+3+...+k)+(k+1)]}
=k(k+1)/2+(k+1)
=(k^2+k+2k+2)/2
=(k^2+3k+2)/2 [:.k^2+3k+2=(k+1)(k+2)]
=(k+1)(k+2)/2//
so, p(k+1) is true.
and so, by the mathematical induction, we know that p(n) is true for every n£N.
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