Need someone to show me the solution Need someone to show me the solution. and tell me how ! 
$$(P÷N) × (N×(N+1)÷2) + N×(1-P) = N×(1-(P÷2)) + (P÷2)$$
 A: \begin{align}
\dfrac{P}{N} \times \dfrac{N(N+1)}2 + N\times (1-P) & = \underbrace{P \times \dfrac{N+1}{2} + N \times (1-P)}_{\text{Cancelling out the $N$ from the first term}}\\
& = \underbrace{\dfrac{PN + P}2 + N - NP}_{\text{$P \times (N+1) = PN + P$ and $N \times (1-P) = N - NP$}}\\
& = \underbrace{\dfrac{PN + P +2N - 2NP}2}_{\text{Take the lcm $2$.}}\\
& = \underbrace{\dfrac{P + 2N -NP}2}_{PN - 2NP = -NP}\\
& = \dfrac{P}2 + \dfrac{2N - NP}2\\
& = \underbrace{N\dfrac{2-P}2 + \dfrac{P}2}_{\text{Factor out $N$ from $2N-NP$}}\\
& = \underbrace{N \left(1 - \dfrac{P}2\right) + \dfrac{P}2}_{\text{Making use of the fact that $\dfrac{2-P}2 = \dfrac22 - \dfrac{P}2 = 1 - \dfrac{P}2$}}
\end{align}
A: $$\begin{align*}\frac{P}{\color{red}{N}}\cdot\frac{\color{red}{N}(N+1)}2+N(1-P)&=\frac{P(N+1)}2+N(1-P)\\
&=\frac{P(N+1)}2+\frac{2N(1-P)}2\\
&=\frac{PN+P+2N-2NP}2\\
&=\frac{P+2N-NP}2\\
&=\frac{P}2+\frac{2N-NP}2\\
&=\frac{P}2+N\left(\frac{2-P}2\right)\\
&=\frac{P}2+N\left(1-\frac{P}2\right)
\end{align*}$$
