Exercise:
Show that, ∞∑n=14(4n−3)(4n+1)=1
Solution: we have 4(4n−3)(4n+1)=A4n−3+B4n+1
So 4A+4B=0&A−3B=4⇔A=1&B=−1
Hence ∞∑n=14(4n−3)(4n+1)=∞∑n=114n−3−∞∑n=114n+1=1+∞∑n=114(n+1)−3−∞∑n=114n+1
∞∑n=14(4n−3)(4n+1)=1+∞∑n=114n+4−3−∞∑n=114n+1=1+∞∑n=114n+1−∞∑n=114n+1=1
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