if n is a positive integer, prove that, 1 cdot 2+2 cdot 3+3 cdot 4+⋯+n cdot (n+1) =(n(n+1)(n+2))/(3)

przesypkai4 2022-07-17 Answered
Discrete math induction problem.
if n is a positive integer, prove that,
1 2 + 2 3 + 3 4 + + n ( n + 1 ) = n ( n + 1 ) ( n + 2 ) 3
Basic step is 1, then assume k is true and we get,
1 2 + 2 3 + 3 4 + + k ( k + 1 ) = k ( k + 1 ) ( k + 2 ) 3
After that I am stuck because the answer key is telling me that the next step is to add ( k + 1 ) ( k + 2 ), but I don't know why we do that. Aren't we supposed to just plugin k + 1 into all k variables. Why are we adding ( k + 1 ) ( k + 2 ) to both sides?
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Answers (2)

Bianca Chung
Answered 2022-07-18 Author has 16 answers
Step 1
So we want to prove that
1 2 + 2 3 + 3 4 + + n ( n + 1 ) = i = 1 n i ( i + 1 ) = n ( n + 1 ) ( n + 2 ) 3 n N
The base case n = 1:
Note that 1 2 = 2 and that 1 2 3 3 = 2. Thus the base case holds.
Suppose that k N such that
i = 1 k i ( i + 1 ) = k ( k + 1 ) ( k + 2 ) 3 . Now consider the case that n = k + 1
i = 1 k + 1 i ( i + 1 ) = ( n + 1 ) ( n + 2 ) + i = 1 k i ( i + 1 )
Step 2
By the induction hypothesis (we claimed that for n = k the formula held) we have:
i = 1 k + 1 i ( i + 1 ) = ( k + 1 ) ( k + 2 ) + k ( k + 1 ) ( k + 2 ) 3 = k 3 + 6 k 2 + 11 k + 6 3 = ( k + 1 ) ( k + 2 ) ( k + 3 ) 3
Therefore, by induction, the formula is true n N
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Cristofer Graves
Answered 2022-07-19 Author has 6 answers
Step 1
Based on hypothesis:
i = 1 k i ( i + 1 ) = 1 3 k ( k + 1 ) ( k + 2 )
it must be shown that:
i = 1 k + 1 i ( i + 1 ) = 1 3 ( k + 1 ) ( k + 2 ) ( k + 3 )
Here k + 1 is 'plugged in' for k.
Step 2
That is the inductionstep. How to do that? Well,
i = 1 k + 1 i ( i + 1 ) = i = 1 k i ( i + 1 ) + ( k + 1 ) ( k + 2 ) = 1 3 k ( k + 1 ) ( k + 2 ) + ( k + 1 ) ( k + 2 )
(in the last equality you are using the hypothesis)
So if you can show that the RHS of this indeed equals:
1 3 ( k + 1 ) ( k + 2 ) ( k + 3 )
then you are ready. Working out the RHS comes to adding ( k + 1 ) ( k + 2 ) to 1 3 k ( k + 1 ) ( k + 2 )
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asked 2022-05-15
G = ( I k | A ) is a generator matrix of C iff H = ( A T | I n k ) is a control matrix of C
I am trying to prove that, ''given a C [n, k, d]-linear code, G = ( I k | A ) is a generator matrix of C iff H = ( A T | I n k ) is a control matrix of C''.
Firstly, I have supposed that G = ( I k | A ) is a generator matrix of C; we name the columns of A as a 1 , a 2 , . . , a n k ; if x = ( x 1 , . . . , x k ) F q k , I can codify this into a word of the code by taking the generator matrix and multiplying it this way: x G = ( x 1 , . . . , x k , x a 1 , . . . , x a n k ) C , where is the dot product of two vectors. I also know the following result:
H is a control matrix of C if the following holds: x C .
So, If I show H ( x 1 , . . . , x k , x a 1 , . . . , x a n k ) T = ( 0 , . . . , 0 ) T , I would have show that x C ; this is very easy to show taking on account how H is constructed.
For the other direction, I would have to show that H x T = ( 0 , . . , 0 ) T has, as a system of equations, as solutions all the words of the code... Here I get stucked.
Secondly, I would have to prove that, assuming H = ( A T | I n k ) is a control matrix of C, G = ( I k | A ) is a generator matrix of C. For this I suppose I would have to solve the system H x T = ( 0 , . . . , 0 ) T (which, by hipothesis I know has as solutions the words of the code), and then show that each of that words can be generated by G, i.e., that, given a code word, I can find a vector in F q k such that the product of that vector by G is the code word firstly given. Nevertheless I am not sure this is the best approach...
Any help, guidance, or anything will be very helpful.
asked 2022-09-06
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For which nonnegative integers n is 2 n + 3 2 n ? Prove your answer.
I know that we first have to inspect for the basis case and so we find that n > 3, since n = 1 , 2 , 3 do not hold but 4 plus does, so n > 3.. But for the Induction step I don't understand how to start it and the textbook solution is unclear, could anyone explain to me what this means?
Textbook solution: For the inductive step assume that P(k) is true. Then, by the inductive hypothesis, 2 ( k + 1 ) + 3 = ( 2 k + 3 ) + 2 < 2 k + 2.. But because k 1 , 2 k + 2 2 k + 2 k = 2 k + 1. This shows that P ( k + 1 ) is true.
asked 2022-06-04
Need a clarification of the proof that the prime ideal space of a distributive bounded lattice is compact
Let L be a bounded distributive lattice, then the prime ideal space I p ( L ) ; τ is compact. τ is the topology whose basis is B = { X b ( X X c ) : b , c L }, with X a = { I I p ( L ) : a I }.
We want to prove that the subbasis S = { X b : b L } { X X c : c L } satisfies Alexander's Lemma. Let U := { X b : b A 0 } { X X c : c A 1 }, a open cover of I p ( L ).
Let J be the ideal generated by A 0 and G the filter generated by A 1 . It is easy to prove that J G . So J G and let a J G; if A 0 and A 1 are both non-empty, there exist b 1 , , b j A 0 and c 1 , , c k A 1 s.t. c 1 c k a b 1 b j , whence X = X 1 = X b 1 X b j ( X X c 1 ) ( X X c k )
What escapes me is why we can write 1 in that way if J G ?

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