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**Unformatted text preview: **this essentially what we’re
trying to prove in the ﬁrst place? To help explain this step a little better, we show how this works
for speciﬁc values of n. We’ve already established P (1) is true, and we now want to show that P (2)
1 Another word for this you may have seen is ‘axiom.’ 574 Sequences and the Binomial Theorem is true. Thus we need to show that a2 = a + (2 − 1)d. Since P (1) is true, we have a1 = a, and by the
deﬁnition of an arithmetic sequence, a2 = a1 + d = a + d = a +(2 − 1)d. So P (2) is true. We now use
the fact that P (2) is true to show that P (3) is true. Using the fact that a2 = a + (2 − 1)d, we show
a3 = a +(3 − 1)d. Since a3 = a2 + d, we get a3 = (a + (2 − 1)d) + d = a + 2d = a + (3 − 1)d, so we have
shown P (3) is true. Similarly, we can use the fact that P (3) is true to show that P (4) is true, and so
forth. In general, if P (k ) is true (i.e., ak = a +(k − 1)d) we set out to show that P (k +1) is true (i.e.,
ak+1 = a + ((k + 1) − 1)d). Assuming ak = a + (k − 1)d, we have by the deﬁnition of an arithmetic
sequence that ak+1 = ak + d so we get ak+1 = (a + (k − 1)d)...

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