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Let $\{a_{n}\}$ be a sequence of real numbers. The backward differences of this sequence are defined recursively as shown next. The first difference $\triangledown a_{n}$ is
$$\triangledown a_{n} = a_{n} − a_{n−1}.$$
The $(k + 1)^{\text{st}}$ difference $\triangledown^{k+1}a_{n}$ is obtained from $\triangledown ^{k} a_{n}$ by
$$\triangledown ^{k+1}a_{n} = \triangledown^{k}a_{n} − \triangledown ^{k}a_{n−1}.$$
Show that $a_{n−2} = a_{n} − 2\triangledown a_{n} + \triangledown^{2}a_{n}.$
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