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Conjugate of a Complex Number

For 2≤ r≤ n...

Question

For $2 \leq r \leq n, (\frac{n}{r}) + 2 (\frac{n}{r - 1}) + (\frac{n}{r - 2})$ is equal to

A

(\frac{n + 1}{r - 1})

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B

2 (\frac{n + 1}{r + 1})

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C

2 (\frac{n + 2}{r})

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D

(\frac{n + 2}{r})

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Solution

The correct option is D $(\frac{n + 2}{r})$
$^{n} C_{r} +^{n} C_{r - 1} +^{n} C_{r - 1} +^{n} C_{r - 2}$
$= (^{n} C_{r} +^{n} C_{r - 1}) + (^{n} C_{r - 1} +^{n} C_{r - 2})$
$=^{n + 1} C_{r} +^{n + 1} C_{r - 1}$
$=^{n + 2} C_{r}$
This comes from the property that
$^{n} C_{r - 1} +^{n} C_{r} =^{n + 1} C_{r}$

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Similar questions

2 \leq r \leq n, (\frac{n}{r}) + 2 (\frac{n}{r - 1}) + (\frac{n}{r + 2}) =

^{n} C_{r - 1} +^{n + 1} C_{r - 1} +^{n + 2} C_{r - 1} + . . . . . . . +^{2 n} C_{r - 1}, =^{2 n + 1} C_{r^{2} - 132} -^{n} C_{r}

, then the value of

r

n \geq r - 1

{^{n} C}_{r - 1} + {^{n + 1} C}_{r - 1} + {^{n + 2} C}_{r - 1} + . . . . + {^{2 n} C}_{r - 1} = {^{2 n + 1} C}_{r^{2} - 182} - {^{n} C}_{r}

then the least possible value of

n

\sum_{r = 0}^{n} r (^{n} C_{r})^{2} = n (^{2 n - 1} C_{n - 1})

\sum_{r = 0}^{n} (^{n} C_{r})^{2} =^{2 n} C_{n}

	Column - I		Column - II
(A)	$\sum_{r = 1}^{n} r .^{n} C_{r}$ is equal to	(P)	$(n + 2) {.2}^{n - 1}$
(B)	$\sum_{r = 1}^{n + 1} r .^{n} C_{r - 1}$ is equal to	(Q)	$(n + 1) .^{2 n} C_{n}$
(C)	$\sum_{r = 0}^{n} (2 r + 1) .^{n} C_{r}$ is equal to	(R)	$(n + 1) {.2}^{n}$
(D)	$\sum_{r = 0}^{n} (2 r + 1) . (^{n} C_{r})^{2}$ is equal to	(S)	$n {.2}^{n - 1}$

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