1-1 -n-1 -1-3 -n-3 -1-n-5 -

The correct option is B Multiplying each term by n! the question reduces to n!/1!(n 1)! + nbsp; 1/3!. nbsp; n!/(n 3)! + nbsp; 1/5!. nbsp; n!/(n 5)! + ...

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Byju's Answer

Standard XI

Mathematics

Proof by mathematical induction

1/1 !n-1 !+1/...

Question

$\frac{1}{1! (n - 1)!}$ + $\frac{1}{3! (n - 3)!}$ + $\frac{1}{(n - 5)!}$ + ....... =

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Solution

The correct option is B

Multiplying each term by n! the question

reduces to

$\frac{n!}{1! (n - 1)!}$ + $\frac{1}{3!}$ . $\frac{n!}{(n - 3)!}$ + $\frac{1}{5!}$ . $\frac{n!}{(n - 5)!}$ + ........

= $^{n} C_{1}$ + $^{n} C_{3}$ + $^{n} C_{5}$ + ........ = $2^{n - 1}$ .

Thus $\frac{1}{1! (n - 1)!}$ + $\frac{1}{3! (n - 3)!}$ + $\frac{1}{5! (n - 5)!}$ +........= $\frac{1}{n!}$ $2^{n - 1}$ .

Suggest Corrections

Similar questions

Q. If

n \in N

n

is even, then

\frac{1}{1. (n - 1)!} + \frac{1}{3! . (n - 3)!} + \frac{1}{5! . (n - 5)!} + . . . . . . + \frac{1}{(n - 1)! 1!} =

Q. If

n \in N

& n is even then

\frac{1}{1! \cdot (n - 1)!}

\frac{1}{3! \cdot (n - 3)!}

\frac{1}{5! \cdot (n - 5)!}

+.....+

\frac{1}{(n - 1)! 1!}

$\frac{1}{1! (n - 1)!}$ + $\frac{1}{3! (n - 3)!}$ + $\frac{1}{5! (n - 5)!}$ + ....... =

Q. Prove that if

n

be an even integer,

\frac{1}{| 1 - | n - 1 - ---- -} + \frac{1}{| 3 - | n - 3 - ---- -} + \frac{1}{| 5 - | n - 5 - ---- -} + . . . . + \frac{1}{| n - 1 - ---- - | 1 -} = \frac{2^{n - 1}}{| n - -}

Q. If

n ϵ N

and

n

is even, then

\frac{1}{1. (n - 1)!} + \frac{1}{3! (n - 3)!} + \frac{1}{5! (n - 5)!} + . . . . + \frac{1}{(n - 1)! 1!}

equals.

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11!(n−1)! + 13!(n−3)! + 1(n−5)! + ....... =

The correct option is B Multiplying each term by n! the question reduces to n!1!(n−1)! + 13!. n!(n−3)! + 15!. n!(n−5)! + ........ = nC1 + nC3 + nC5 + ........ = 2n−1. Thus 11!(n−1)! + 13!(n−3)! + 15!(n−5)! +........= 1n!2n−1.

$\frac{1}{1! (n - 1)!}$ + $\frac{1}{3! (n - 3)!}$ + $\frac{1}{(n - 5)!}$ + ....... =