Prove that two +ive integers n, r cannot be found such that $^{n} C_{r},^{n} C_{r + 1},^{n} C_{r + 2} a n d^{n} C_{r + 3}$ are in A.P.

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Solution

The first three being in A.P implies that
$2^{n} C_{r + 1} =^{n} C_{r} +^{n} C_{r + 2}$
The given terms are the coefficients of $T_{r + 1}, T_{r + 2}, T_{r + 3}, T_{r + 4}$ . The first being A.P implies $2 T_{r + 2} = T_{r + 1} + T_{r + 3}$ or $2 = \frac{T_{r + 1}}{T_{r + 2}} + \frac{T_{r + 3}}{T_{r + 2}}$
or $2 = \frac{1}{^{n} C_{r + 1} / n C_{r}} + \frac{^{n} C_{r + 2}}{^{n} C_{r + 1}}$ ...(1)
Now $\frac{^{n} C_{r + 1}}{^{n} C_{r}} = \frac{n!}{(n - r - 1)! (r + 1)!} - \frac{(n - r)! r!}{n!} = \frac{n - r}{r + 1}$ ...(2)
replacing r by r + 1, we have
$\frac{^{n} C_{r + 2}}{^{n} C_{r + 1}} = \frac{n - r - 1}{r + 2}$ .....(3)
Hence from (1) by the help of (2) and (3) , we have
$2 = \frac{r + 1}{n - r} = \frac{n - r - 1}{r + 2}$
or (n - r )(2r + 4) = (r $^{2}$ + 3r + 2) + (n - r) $^{2}$ - (n - r)
or (n - r) $^{2}$ - (n - r)(2r + 5) + (r + 1)(r + 2) = 0 .....(4)
Above is the condition for first three terms to be in A.P .Again replacing r by r + 1, we shall get the condition for the last three terms to be in A.P as
(n - - r - 1) $^{2}$ - (n - r - 1)(2x + 7) + (r + 2)(r + 3) = 0
or (n - r) $^{2}$ + 1 - 2(n - r) - (n - r)(2r + 7) + (2r + 7) + $r^{2}$ + 5r + 6 = 0
or (n - r) $^{2}$ - (n - r )(2r + 9) + ( $r^{2}$ + 7r + 13) = 0 ....(5)
comparing (4) and (5) , we have
$2 = \frac{2 r + 5}{2 r + 9} = \frac{r^{2} + 3 r + 2}{r^{2} + 7 r + 13}$
1st and 2nd give 9 = 5 and 1st and 3rd give -ive value for r. Both are not possible. Hence the given four terms cannot be in A.P.

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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 n and r are two positive integers such that

n \geq r

, then

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

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.

Q. If

n

and

r

are positive integers such that

r < n

, then

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

Q. Prove that :

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

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Prove that two +ive integers n, r cannot be found such that nCr,nCr+1,nCr+2andnCr+3 are in A.P.

Prove that two +ive integers n, r cannot be found such that $^{n} C_{r},^{n} C_{r + 1},^{n} C_{r + 2} a n d^{n} C_{r + 3}$ are in A.P.