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+)
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]=
[14] [2] = 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+)
(28+) - (27+) =
(28)(29)/ 2 - (27)(28)/2 =
(28) [ (29) - (27) ] / 2 =
(28/2) (2) = 28 = (7)(8)/ 2 = (7+) ...so k = 7
I AN NOT GETTING WHAT YOU HAVE GIVEN PLEASE EXPLAIN YOUR ANS TO ME MORE BRIEFLY
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+)
$$\small{\text{$
\begin{array}{rcl}
(28+)-(27+) = (k+) = 28 &=& S_k \qquad (28+) = S_{28} \quad (27+)=S_{27}\\\\
S_{28}-S_{27} = t_{28} = 28&=&S_k\\\\
S_k &=& t_1 \cdot \dbinom{k}{1} + d\cdot \dbinom{k}{2} \qquad t_1=d=1\\\\
S_k &=& \dbinom{k}{1} + \dbinom{k}{2} = \dbinom{k+1}{2} = 28\\\\ \\
\dfrac{k(k+1)}{2} &=& 28 \\\\
k(k+1) &=& 56\\\\
k^2+k-56&=&0\\\\
\Rightarrow k &=& \dfrac{-1+15}{2} = 7\\\\
\mathbf{S_k = (k+) }& \mathbf{=} & \mathbf{(7+)}\\\\
(7+)=1+2+3+4+5+6+7&=&28 \\
\end{array}
$}}$$
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]=
[14] [2] = 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+)