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Need someone to show me the solution. and tell me how ! $$(P÷N) × (N×(N+1)÷2) + N×(1-P) = N×(1-(P÷2)) + (P÷2)$$

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up vote 1 down vote accepted

$$\begin{align*}\frac{P}{\color{red}{N}}\cdot\frac{\color{red}{N}(N+1)}2+N(1-P)&=\frac{P(N+1)}2+N(1-P)\\ &=\frac{P(N+1)}2+\frac{2N(1-P)}2\\ &=\frac{PN+P+2N-2NP}2\\ &=\frac{P+2N-NP}2\\ &=\frac{P}2+\frac{2N-NP}2\\ &=\frac{P}2+N\left(\frac{2-P}2\right)\\ &=\frac{P}2+N\left(1-\frac{P}2\right) \end{align*}$$

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\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}

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