mathematic inductionProve,by mathematic inductiona)1*2+2*3+3*4+.+n(n+1)=1/3*n(n+1)(n+2)b)hence or otherwise find the value of 1+(1+2)+(1+2+3)+...+(1+2+3+.+100)

mathematic inductionProve,by mathematic inductiona)1*2+2*3+3*4+.+n(n+1)=1/3*n(n+1)(n+2)b)hence or otherwise find the value of 1+(1+2)+(1+2+3)+...+(1+2+3+.+100)
mathematic induction
Prove,by mathematic induction
a)1*2+2*3+3*4+.+n(n+1)=1/3*n(n+1)(n+2)
b)hence or otherwise find the value of
1+(1+2)+(1+2+3)+...+(1+2+3+.+100)

mathematic inductionProve,by mathematic inductiona)1*2+2*3+3*4+.+n(n+1)=1/3*n(n+1)(n+2)b)hence or otherwise find the value of 1+(1+2)+(1+2+3)+...+(1+2+3+.+100)
a)
first:1*2=1/3*1*2*3=2
2nd:if 1*2+2*3+3*4+.+n(n+1)=1/3*n(n+1)(n+2) then:
1*2+2*3+3*4+.+n(n+1)+(n+1)(n+2)=1/3*n(n+1)(n+2)+(n+1)(n+2)=(n+1)(n+2)(n/3+1)=1/3(n+1)(n+2)(n+3)
so:1*2+2*3+3*4+.+n(n+1)=1/3*n(n+1)(n+2)
b)1+(1+2)+(1+2+3)+...+(1+2+3+.+100)
=1*2/2+2*3/2+3*4/2+.+100(100+1)/2=1/3*100*101*102/2=171700

let P(n) be the statement that 1*2+2*3+3*4+...........+n(n+1)=1/3*n(n+1)(n+2)
consider P(1)
left hand side=1*2=2
right hand side=1/3*1*2*3=2=left hand side
so P(1) is true
Suppose ...

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let P(n) be the statement that 1*2+2*3+3*4+...........+n(n+1)=1/3*n(n+1)(n+2)
consider P(1)
left hand side=1*2=2
right hand side=1/3*1*2*3=2=left hand side
so P(1) is true
Suppose P(k) is true for some integer k
i.e. 1*2+2*3+3*4+...........+k(k+1)=1/3*k(k+1)(k+2)
consider P(k+1)
left hand side=1*2+2*3+3*4+...........+k(k+1)+(k+1)(k+2)=1/3*k(k+1)(k+2)+(k+1)(k+2)=1/3(k+1)(k+2)(k+3)=right hand side
therefore P(k) is true implies that P(k+1) is true
since P(1) is true,
by mathematical induction,
P(n) is true for all integers n (proven)
b)method is the same as given by wuhchau
1+(1+2)+(1+2+3)+...+(1+2+3+......+100)
=1*2/2+2*3/2+3*4/2+...........+100(100+1)/2=1/3*100*101*102/2=171700

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a)
we know that
1+2+3+...+n=1/2*n*(n+1)
1^2+2^2+3^2+...+n^2=1/6*n*(n+1)*(2n+1)
n(n+1)=n^2+n
hence
1*2+2*3+3*4+...........+n(n+1)
=(1^2+2^2+3^2+...+n^2)+(1+2+3+...+n)
=...

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a)
we know that
1+2+3+...+n=1/2*n*(n+1)
1^2+2^2+3^2+...+n^2=1/6*n*(n+1)*(2n+1)
n(n+1)=n^2+n
hence
1*2+2*3+3*4+...........+n(n+1)
=(1^2+2^2+3^2+...+n^2)+(1+2+3+...+n)
=1/6*n*(n+1)*(2n+1)+1/2*n*(n+1)
=1/6*n*(n+1)*(2n+1+3)
=1/3*n*(n+1)*(n+2)
b)
we know that 1+2+3+...+n=1/2*n*(n+1)
hence
1+(1+2)+(1+2+3)+...+(1+2+3+......+100)
=1/2*1*2+1/2*2*3+1/2*3*4+...+1/2*100*101
=1/2*(1*2+2*3+3*4+...+100*101)
=1/2*1/3*100*101*102
=171700

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=171700