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# If we use (x+) to indicate the following sum

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If we use (x+) to indicate the following sum
1 + 2 + 3 + ... + x
then find the value of k in the following equation
(28+) - (27+) = (k+)

Aug 12, 2015

#4
+10

The sum of the first "k" positive integers is given by :   (k)(k + 1)/2

So  (x +)   where x =  28  is just  (28)(29) /2     and  (x+) where x = 27  is just (27)(28)/2

So we have

(28)(29)/2 -  (270(28)/2        ..... factor out 28/2

[28/2] [ 29 - 27]=

   = 28

And notice that :

1 + 2 + 3 + 4 + 5 + 6 + 7  = 28

So this equals the sum of the first seven positive integers  =  (7)(8)/2  = 56/2 = (7+)   Aug 12, 2015

#1
+5

(28+) - (27+) =

(28)(29)/ 2 - (27)(28)/2  =

(28) [ (29) - (27) ] / 2 =

(28/2) (2) = 28  =  (7)(8)/ 2 =   (7+)   ...so k = 7   Aug 12, 2015
#2
0

I AN NOT GETTING WHAT YOU HAVE GIVEN PLEASE EXPLAIN YOUR ANS TO ME MORE BRIEFLY Aug 12, 2015
#3
+10

If we use (x+) to indicate the following sum
1 + 2 + 3 + ... + x
then find the value of k in the following equation
(28+) - (27+) = (k+)

$$\small{\text{ \begin{array}{rcl} (28+)-(27+) = (k+) = 28 &=& S_k \qquad (28+) = S_{28} \quad (27+)=S_{27}\\\\ S_{28}-S_{27} = t_{28} = 28&=&S_k\\\\ S_k &=& t_1 \cdot \dbinom{k}{1} + d\cdot \dbinom{k}{2} \qquad t_1=d=1\\\\ S_k &=& \dbinom{k}{1} + \dbinom{k}{2} = \dbinom{k+1}{2} = 28\\\\ \\ \dfrac{k(k+1)}{2} &=& 28 \\\\ k(k+1) &=& 56\\\\ k^2+k-56&=&0\\\\ \Rightarrow k &=& \dfrac{-1+15}{2} = 7\\\\ \mathbf{S_k = (k+) }& \mathbf{=} & \mathbf{(7+)}\\\\ (7+)=1+2+3+4+5+6+7&=&28 \\ \end{array} }}$$ Aug 12, 2015
#4
+10

The sum of the first "k" positive integers is given by :   (k)(k + 1)/2

So  (x +)   where x =  28  is just  (28)(29) /2     and  (x+) where x = 27  is just (27)(28)/2

So we have

(28)(29)/2 -  (270(28)/2        ..... factor out 28/2

[28/2] [ 29 - 27]=

   = 28

And notice that :

1 + 2 + 3 + 4 + 5 + 6 + 7  = 28

So this equals the sum of the first seven positive integers  =  (7)(8)/2  = 56/2 = (7+)   CPhill Aug 12, 2015