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  title: '"I think it would be worthwhile to explicitly ca..."',
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  anchorContext: 'We'll do an example to build our intuition before giving the proper definition of the principle\\. We'll provide yet another proof that\n$$ 1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}$$\nfor all integers $n \\ge 1$\\. First, let's check the base case, where $n=1$:\n$$ 1 = \\frac{1(1+1)}{2} = \\frac{2}{2} = 1.$$\nThis is \\(fairly obviously\\) true, so we can move forward with the inductive step\\. The inductive step includes an assumption, namely that the statement is true for some integer $k$ that is larger than the base integer\\. For our example, if $k\\ge1$, we assume that\n$$1 + 2 + \\cdots + k = \\frac{k(k+1)}{2}$$\nand try to prove that\n$$ 1 + 2 + \\cdots + k + (k+1) = \\frac{(k+1)([k+1]+1)}{2}.$$\nTake our assumption and add $k+1$ to both sides:\n$$1+2+\\cdots + k + (k+1) = \\frac{k(k+1)}{2} + k + 1.$$\nNow the left\\-hand sides of what we know and what we want are the same\\. Hopefully the right\\-hand side will shake out to be the same\\. Get common denominators so that the right\\-hand side of the above equation is\n$$\\frac{k(k+1)}{2} + \\frac{2(k+1)}{2} = \\frac{(k+2)(k+1)}{2} = \\frac{(k+1)([k+1]+1)}{2}.$$\nTherefore,\n$$ 1 + 2 + \\cdots + k + (k+1) = \\frac{(k+1)([k+1]+1)}{2}$$\nas desired\\.',
  anchorText: 'and try to prove that',
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  text: 'I think it would be worthwhile to explicitly call out that what's happening here is that we're replacing $n$ in the original equation with $k+1$.',
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