Prove that -r - 0n - - 2r - nCrr - 2Cr - -1-n - 1- n even1-n - 2- n odd

As in last part (A) we have ^nC_r^r + 2C_r = 2(n + 2) #160; #160;(n + 1) ^n + 2C_r + 2 ∑ = 2(n + 2) #160; #160;( ...

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Byju's Answer

Standard XII

Mathematics

Algebraic Expression

prove that ...

Question

prove that
$\sum_{r = 0}^{n} \cdot (- 2)^{r} \cdot \frac{^{n} C_{r}}{^{r + 2} C_{r}} = [_{1 / (n + 2), n o d d}^{1 / (n + 1), n e v e n}$

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Solution

As in last part (A) we have
$\frac{^{n} C_{r}}{^{r + 2} C_{r}} = \frac{2}{(n + 2) (n + 1)}^{n + 2} C_{r + 2}$
$\sum = \frac{2}{(n + 2) (n + 1)} [\sum (- 2)^{r} \cdot^{n + 2} C_{r + 2}]$
Now put r + 2 = s $∴$ r = s - 2
$∴ (- 2)^{r} = (- 2)^{s} (- 2)^{- 2} = \frac{1}{4} (- 2)^{s}$
$\sum = \frac{1}{2 (n + 2) (n + 1)} [\sum_{s = 0}^{n + 2} (- 2)^{s}^{n + 2} C_{s} - [(- 2)^{0}^{n + 2} C_{0} + (- 2)^{1}^{n + 2} C_{1}]]$
$= \frac{1}{2 (n + 2) (n + 1)} [(1 - 2)^{n + 2} - 1 + 2 (n + 2)]$
$= \frac{1}{2 (n + 2) (n + 1)} [(- 1)^{n + 2} + 2 n + 3]$
Now if n is even then $N^{r}$ = 2n + 4 = 2(n + 2)
When n is odd then $N^{r}$ = 2n + 2 = 2(n + 1)etc.

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Similar questions

Q. Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
(a)

\frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \frac{n - r + 1}{r}

(b) n ·_{^{n −}}_¹C_r_{− 1} = (n − r + 1) ⁿC_r_{− 1}
(c)

\frac{{}^{n}C_{r}}{{}^{n - 1}C_{r - 1}} = \frac{n}{r}

(iv) ⁿC_r + 2 · ⁿC_r_{− 1} + ⁿC_r_{− 2} = ^{n +}²C_r.

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. The value of

n \sum r = 0 (- 1)^{r} \frac{^{n} C_{r}}{^{r + 2} C_{r}}

is equal to

Q. Assertion :For

n \geq 1

, let

C_{r} = (\begin{matrix} n r \end{matrix})

0 \leq r \leq n

{lim}_{x \to \infty} \sum_{r = 0}^{n} \frac{C_{r}}{(r + 3) n^{r}} = e - 2

Reason:

\sum_{r = 0}^{n} \frac{C_{r}}{(r + 3) n^{r}} = \int_{0}^{1} {(1 + \frac{x}{n})}^{n} x^{2} d x

Q. If

a_{n} = n \sum r = 0 \frac{1}{{^{n} C}_{r}}

then

n \sum r = 0 \frac{r}{{^{n} C}_{r}}

equals

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prove that ∑nr=0⋅(−2)r⋅nCrr+2Cr=[1/(n+1),neven1/(n+2),nodd

prove that
$\sum_{r = 0}^{n} \cdot (- 2)^{r} \cdot \frac{^{n} C_{r}}{^{r + 2} C_{r}} = [_{1 / (n + 2), n o d d}^{1 / (n + 1), n e v e n}$