$^{n} C_{0} +^{n + 1} C_{1} +^{n + 2} C_{2} + . . . . +^{n + r} C_{r} =$

^{n + r + 1} C_{r}

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^{n + r - 1} C_{r}

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n

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^{n + r + 2} C_{r}

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Solution

The correct option is A $^{n + r + 1} C_{r}$
$^{n} C_{0} +^{n + 1} C_{1} + . . .^{n + r} C_{r}$
$= (^{n} C_{0} +^{n + 1} C_{0} +^{n + 1} C_{1} + . . .^{n + r} C_{r}) -^{n + 1} C_{0}$
$= (^{n} C_{0} +^{n + 2} C_{1} +^{n + 2} C_{2} + . . .^{n + r} C_{r}) -^{n + 1} C_{0}$ .... $[∵^{n + 1} C_{0} +^{n + 1} C_{1} =^{n + 2} C_{1}]$
$= (^{n} C_{0} +^{n + 3} C_{2} +^{n + 3} C_{3} + . . .^{n + r} C_{r}) -^{n + 1} C_{0}$ ... $[∵^{n + 2} C_{1} +^{n + 2} C_{2} =^{n + 3} C_{2}]$
:
:
$= (^{n} C_{0} +^{n + r} C_{r - 1} +^{n + r} C_{r}) -^{n + 1} C_{0}$
$= (^{n} C_{0} +^{n + r + 1} C_{r}) -^{n + 1} C_{0}$ ..... $[∵^{n + r} C_{r - 1} +^{n + r} C_{r} =^{n + r + 1} C_{r}]$
$= 1 +^{n + r + 1} C_{r} - 1$
$=^{n + r + 1} C_{r}$ .

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Similar questions

Q. Prove that

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

For $1 \leq r \leq n$ , the value of $ {}^nC_r +^{{}^{n-1}C_r}+^{{}^{n-2}C_r}.....+...+....+^{{}^rC_r} \space is : $

Q. Let n and r be two positive integers such that

n \geq r + 2

. Suppose

Δ (n, r) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{r} & ^{n} C_{r + 1} & ^{n} C_{r + 2}^{n + 1} C_{r} & ^{n + 1} C_{r + 1} & ^{n + 1} C_{r + 2}^{n + 2} C_{r} & ^{n + 2} C_{r + 1} & ^{n + 2} C_{r + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Show that

Δ (n, r) = \frac{^{n + 2} C_{3}}{^{n + 2} C_{3}} Δ (n - 2, r - 1)

Hence or otherwise,

Let r and n be positive integers such that $1 \leq r \leq n .$ Then prove the following :

(i) $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
(ii) $n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$
(iii) $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$
(iv) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

Q. If

(2 \leq r \leq n),

then

^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}

is equal to

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nC0+n+1C1+n+2C2+....+n+rCr=

$^{n} C_{0} +^{n + 1} C_{1} +^{n + 2} C_{2} + . . . . +^{n + r} C_{r} =$