{
  "translations" :{
    "en": {
      "sc1": "We will use expansion to guess the closed form solution for the recurrence $\\mathbf{T}(n) = \\mathbf{T}(n-1) + 1$ for $n > 1$; $\\mathbf{T}(0) = \\mathbf{T}(1) = 0$.",
      "sc2.1": "When we expand the recurrence for a problem of size $n$, we replace the problem of size $n$ with $1$ unit of work plus a subproblem of size $n-1$:",
      "sc2.2": " $\\mathbf{T}(n) = 1 + \\mathbf{T}(n-1)$",
      "sc3.1": "We replace a problem of size $n-1$ with $1$ unit of work plus a subproblem of size $n-2$",
      "sc3.2": "<br> $\\mathbf{T}(n) = 1 + (1 + \\mathbf{T}(n-2))$",
      "sc4.1": "We replace a problem of size $n-2$ with $1$ unit of work plus a subproblem of size $n-3$",
      "sc4.2": "<br> $\\mathbf{T}(n) = 1 + (1 + (1 + \\mathbf{T}(n-3)))$",
      "sc5.1": "We replace a problem of size $n-3$ with $1$ unit of work plus a subproblem of size $n-4$",
      "sc5.2": "<br> $\\mathbf{T}(n) = 1 + (1 + (1 + (1 + \\mathbf{T}(n-4)))$",
      "sc6.1": "This pattern will continue till we reach a subproblem of size $1$",
      "sc6.2": "<br> $\\mathbf{T}(n) = 1 + (1 + (1 + (1 + (1 + (...))))$",
      "sc7": "Thus, the closed form solution of $\\mathbf{T}(n) = \\mathbf{T}(n-1) + 1$ can be modeled by the summation $\\displaystyle\\sum_{i=1}^{n}1$",
      "sc8": "This gives us a closed form solution of $\\mathbf{T}(n) = \\mathbf{T}(n-1) + 1 = n$"
    }
  }
}