Question
C01 + C23 + C45 + C67 +..........=
Answer: Option C
:
C
Putting the values of C0,C2,C4........., we get
= 1 + n(n−1)3.2! + n(n−1)(n−2)(n−3)5.4! + .........
= 1n+1[(n+1) + (n+1)n(n−1)3! + (n+1)n(n−1)(n−2)(n−3)5!+ ..........]
Put n+1 = N
= 1N[N + N(N−1)(N−2)3! + N(N−1)(N−2)(N−3)(N−4)5! + .........]
= 1N{NC1 +NC3+NC5+..........}
= 1N{2N−1} = 2nn+1 { ∵ N = n+1}
Trick:Put n = 1, then S1 = 1C01 = 11 = 1
At n = 2, S2 = 2C01 + 2C23 = 1 + 13 = 43
Also (c) ⇒ S1 = 1, S2 = 43
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:
C
Putting the values of C0,C2,C4........., we get
= 1 + n(n−1)3.2! + n(n−1)(n−2)(n−3)5.4! + .........
= 1n+1[(n+1) + (n+1)n(n−1)3! + (n+1)n(n−1)(n−2)(n−3)5!+ ..........]
Put n+1 = N
= 1N[N + N(N−1)(N−2)3! + N(N−1)(N−2)(N−3)(N−4)5! + .........]
= 1N{NC1 +NC3+NC5+..........}
= 1N{2N−1} = 2nn+1 { ∵ N = n+1}
Trick:Put n = 1, then S1 = 1C01 = 11 = 1
At n = 2, S2 = 2C01 + 2C23 = 1 + 13 = 43
Also (c) ⇒ S1 = 1, S2 = 43
Was this answer helpful ?
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