a)
1+1+...+1=i=1∑n1=n
1⋅21+2⋅31+...+n(n+1)1=i=1∑nn(n+1)1=n+1n
b) Let P(n) be the proposition that 1+1+...+1=i=1∑n1, is n. We must prove that P(n) is true for n=1,2,3,...
BASIS STEP: P(1) is true, because 1=i=1∑11=1.
INDUCTIVE STEP: For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that
1+1+...+1=i=1∑k1=k Under this assumption, it must be shown that P(k+1) is true, namely, that
1+1+...+1=i=1∑k+11=k+1is also true.
When we add 1 to both sides of the equation in P(k), we obtain
i=1∑k1+1=i=1∑k+11=k+1This last equation shows that P(k+1) is true under the assumption that P(k) is true. This completes the inductive step.
We have completed the basis step and the inductive step, so by mathematical induction we know that P(n) is true for all positive integers n. That is, we have proven that
1+1+...+1=i=1∑n1=nfor all positive integers n.
Let P(n) be the proposition that 1⋅21+2⋅31+...+n(n+1)1=i=1∑nn(n+1)1 is n+1n. We must prove that P(n) is true for n=1,2,3,...
BASIS STEP: P(1) is true, because 1⋅21=i=1∑1n(n+1)1=1+11=21.
INDUCTIVE STEP: For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that
1⋅21+2⋅31+...+k(k+1)1=i=1∑kk(k+1)1=k+1k Under this assumption, it must be shown that P(k+1) is true, namely, that
1⋅21+2⋅31+...+k(k+1)1+(k+1)(k+1+1)1=i=1∑k+1k(k+1)1=k+1+1k+1 is also true.
When we add (k+1)(k+1+1)1 to both sides of the equation in P(k), we obtain
i=1∑kk(k+1)1+(k+1)(k+1+1)1=i=1∑k+1k(k+1)1=i=1∑k+1k(k+1)1=k+1k+(k+1)(k+1+1)1
=(k+1)(k+2)k(k+2)+1=(k+1)(k+2)(k+1)2
=k+2k+1=k+1+1k+1
This last equation shows that P(k+1) is true under the assumption that P(k) is true. This completes the inductive step.
We have completed the basis step and the inductive step, so by mathematical induction we know that P(n) is true for all positive integers n. That is, we have proven that
1⋅21+2⋅31+...+n(n+1)1=i=1∑nn(n+1)1=n+1nfor all positive integers n.
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