{
  "translations" :{
    "en": {
      "sc1": "We will use expansion to guess the closed form solution for the recurrence $\\mathbf{T}(n) = \\mathbf{T}(n-1) + n\\ \\mbox{for}\\ n > 1; \\mathbf{T}(1) = 1$.",
      "sc2.1": "We replace a problem of size $n$ with $n$ units of work plus a subproblem of size $n-1$.",
      "sc2.2": "<br> $\\mathbf{T}(n) = n + \\mathbf{T}(n-1)$.",
      "sc3.1": "We replace a problem of size $n-1$ with $n-1$ units of work plus a subproblem of size $n-2$.",
      "sc3.2": "<br> $\\mathbf{T}(n) = n + (n-1 + \\mathbf{T}(n-2))$.",
      "sc4.1": "We replace a problem of size $n-2$ with $n-2$ units of work plus a subproblem of size $n-3$.",
      "sc4.2": "<br> $\\mathbf{T}(n) = n + (n-1 + (n-2 + \\mathbf{T}(n-3)))$.",
      "sc5.1": "We replace a problem of size $n-3$ with $n-3$ units of work plus a subproblem of size $n-4$.",
      "sc5.2": "<br> $\\mathbf{T}(n) = n + (n-1 + (n-2 + (n-3 + \\mathbf{T}(n-4)))$.",
      "sc6.1": "This pattern will continue till we reach a subproblem of size $1$, which we know costs 1 unit of work.",
      "sc6.2": "<br> $\\mathbf{T}(n) = n + (n-1 + (n-2 + (n-3 + (n-4 + (... + 1))))$.",
      "sc7": "Thus, the closed form solution of $\\mathbf{T}(n) = \\mathbf{T}(n-1) + n$ can be modeled by the summation $\\displaystyle\\sum_{i=1}^{n}i$.",
      "sc8": "Since we already know that $\\displaystyle\\sum_{i=1}^{n}i = \\frac{n(n+1)}{2}$, we have the closed form solution of $\\mathbf{T}(n) = \\mathbf{T}(n-1) + n = \\frac{n(n+1)}{2}$."
    }
  }
}