Melody

avatar
Nombre de usuarioMelody
Puntuación110100
Membership
Stats
Preguntas 878
Respuestas 29497

 #11
avatar+110100 
+2

a)

\(\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\ \text{by counting the number of ordered triples (a, b) of positive integers}, \\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\\ Where \;\;\;n\ge3\;\;\;and \;\;\;n\in Z\)

 

I am going to prove this using Induction.

 

STEP 1

Prove true for n=3

\(LHS=n^3 \\ LHS=3^3 \\ LHS=27\\ RHS= n + 3n(n - 1) + 6 \binom{n}{3}\\ RHS= 3 + 3*3(3 - 1) + 6 \binom{3}{3}\\ RHS= 3 + 18 + 6 \\ RHS=27\\ RHS=LHS\\ \text{So the statement is true for n=3}\)

 

STEP 2

If true for n=k prove true for n=k+1

so we can assume

\(k^3=k+3k(k-1)+6\binom{k}{3}\\ k^3=k+3k^2-3k+6\binom{k}{3}\\ k^3=3k^2-2k+6\binom{k}{3}\\\)

Now I need to prove it will be true for n=k+1

i.e.  Prove

\((k+1)^3=(k+1)+3(k+1)(k+1-1)+6\binom{k+1}{3}\\ (k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\ \qquad \qquad \qquad \text{An aside:}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)!}{3!(k+1-3)!}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{k!(k+1)}{3!(k-3)!(k-2)}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k-2)+3}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\\\)

 

\(LHS=(k+1)^3\\ LHS=k^3+3k^2+3k+1\)

 

\(RHS=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\ RHS=(k+1)+3(k+1)(k)+6\left[ \binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\right]\\ RHS=k+1+3k^2+3k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=1+3k^2+4k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=3k^2-2k+1+6 \binom{k}{3}+6k+\frac{6*3}{k-2}*\binom{k}{3}\\ substituting\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{k!}{3!(k-3)!}\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{(k-3)!(k-2)(k-1)k}{3!(k-3)!}\\ RHS=k^3+6k+1+\frac{6*3}{1}*\frac{(k-1)k}{3!}\\ RHS=k^3+6k+1+3k^2-3k\\ RHS=k^3+3k^2+3k+1\\ RHS=LHS\\~\\ \text{So if it is true for n=k then it will be true also for n=k+1} \)

 

Step 3

Since it is true for n=3 it must be true for n=4, n=5, n=6,  ....

Hence it must be true for all integer n where n greater or equal to 3.

 

 

 

 

 

LaTex:

\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\
 \text{by counting the number of ordered triples (a, b) of positive integers}, 
\\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\\
Where \;\;\;n\ge3\;\;\;and \;\;\;n\in Z

 

LHS=n^3        \\                              
LHS=3^3 \\
LHS=27\\
RHS= n + 3n(n - 1) + 6 \binom{n}{3}\\
RHS= 3 + 3*3(3 - 1) + 6 \binom{3}{3}\\
RHS= 3 + 18 + 6 \\
RHS=27\\
RHS=LHS\\
\text{So the statement is true for n=3}

 

k^3=k+3k(k-1)+6\binom{k}{3}\\
k^3=k+3k^2-3k+6\binom{k}{3}\\
k^3=3k^2-2k+6\binom{k}{3}\\

 

(k+1)^3=(k+1)+3(k+1)(k+1-1)+6\binom{k+1}{3}\\
(k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\
\qquad \qquad \qquad \text{An aside:}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)!}{3!(k+1-3)!}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{k!(k+1)}{3!(k-3)!(k-2)}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k-2)+3}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\\

 

LHS=(k+1)^3\\
LHS=k^3+3k^2+3k+1

 

RHS=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\
RHS=(k+1)+3(k+1)(k)+6\left[ \binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\right]\\
RHS=k+1+3k^2+3k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=1+3k^2+4k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=3k^2-2k+1+6 \binom{k}{3}+6k+\frac{6*3}{k-2}*\binom{k}{3}\\
substituting\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{k!}{3!(k-3)!}\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{(k-3)!(k-2)(k-1)k}{3!(k-3)!}\\
RHS=k^3+6k+1+\frac{6*3}{1}*\frac{(k-1)k}{3!}\\
RHS=k^3+6k+1+3k^2-3k\\
RHS=k^3+3k^2+3k+1\\
RHS=LHS\\~\\
\text{So if it is true for n=k then it will be true also for n=k+1}

 

 


 

5 ago. 2020 16:56
 #5
avatar+110100 
+3

Thanks Heureka, 

Here is my contribution. 

It is not much different from yours, but the explanation is maybe a little more different.

 

 

If a 7-digit number 13ab45c is divisible by 792, what is b?

 

\(792=8*9*11\)

 

For this number to be dividable by 8, the last 3 digits must be divisible by 8

456/8=57  no other digits will work so  c=6

 

\(13ab45c\\ becomes\\ 13ab456\)

 

Now a number is divisible by 11 if

| (sum of even digits) - (sum of odd digits) | = multiple of 11

 

Sum of odd digits = 1+a+4+6 = 11+a

Sum of even digits = 3+b+5 = 8+b

absolute value of difference = | 11+a - 8-b | = |3+a-b |  

so 3+a+b can be negative but it must be a multiple of 11

3+a-b = 11k    where k is an integer.

\(-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ a=8+b\;\;or \;\; a=-3+b\)

 

 

Now for a number to be divisible by 9, the sum of the digits must be a multiple of 9

so

\(1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad =27+2b\\ \quad b=0\;\;or\;\;9\\ b=9 \;\;doesn't \;\;work\\ so\;\; b=0,\;\;a=8 \; \text{is a possible solution}\\ ~\\ 1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad =16+2b\\ \quad b=1,\;\;a=-2 \;\text{which is invalid}\)

 

If b=9 then a=17 or 6     17 is not good so

If b=9  a=-3+9=6

If b=0  a=8+0=8

 

So the number could be

1380456   or

1369456

 

So the number is

\(13ab45c= 13ab456 =1380456\\ a=8\\ b=0\\ c=6\)

 

 

 

 

 

LaTex:

-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ 
a=8+b\;\;or \;\; a=-3+b

 

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad  =27+2b\\ \quad b=0\;\;or\;\;9\\
b=9 \;\;doesn't \;\;work\\
so\;\;   b=0,\;\;a=8 \; \text{is a possible solution}\\
~\\

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad  =16+2b\\
 \quad b=1,\;\;a=-2 \;\text{which is invalid}

4 ago. 2020