Question
Prove that
1−2n+2n(2n−1)2!−2n(2n−1)(2n−1)3!+...+(−1)n−12n(n−1)...(n+2)(n−1)
=(−1)n+1(2n)2(n!)2, where n is a + ive integer.
1−2n+2n(2n−1)2!−2n(2n−1)(2n−1)3!+...+(−1)n−12n(n−1)...(n+2)(n−1)
=(−1)n+1(2n)2(n!)2, where n is a + ive integer.
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Solution
L.H.S.=S=1−2nC1+2nC2−......+(−1)n−12nCn−1 n terms Now we know that 2nC0=1=2nC2n2nC1=2nC2n−1 etc.∴2S=(2nC0+2nC2n)−(2nC1+2nC2n−1)+.....+(−1)n−1(2nCn−1+2nCn+1) 2S=2nC0−2nC1+2nC2−......−2nC2n−1+2nC2n 2n terms Add (−1)n 2nCn to both sides to make (2n+1) terms 2S+(−1)n2nCn=C0−C1+C2−+C2n=0[ on putting x = -1 in the expansion of (1+x)2n which contains (2n + 1) terms ] ∴S=−12(−1)n2nCn=(−1)n+1(2n)!2(n!)2=R.H.S.
L.H.S.
=S=1−2nC1+2nC2−......+(−1)n−12nCn−1 n terms
Now we know that 2nC0=1=2nC2n
2nC1=2nC2n−1 etc.
∴2S=(2nC0+2nC2n)−(2nC1+2nC2n−1)+.....+(−1)n−1(2nCn−1+2nCn+1)
2S=2nC0−2nC1+2nC2−......−2nC2n−1+2nC2n 2n terms
Add (−1)n 2nCn to both sides to make (2n+1) terms
2S+(−1)n2nCn=C0−C1+C2−+C2n=0
[ on putting x = -1 in the expansion of (1+x)2n which contains (2n + 1) terms ]
∴S=−12(−1)n2nCn=(−1)n+1(2n)!2(n!)2=R.H.S.
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