For 2 r n , ( n r ) + 2 ( n r - 1 ) + ( n r - 2 ) = [2000]

Question

For $2 \leq r \leq n, (\binom{n}{r}) + 2 (\binom{n}{r - 1}) + (\binom{n}{r - 2}) =$ [2000]

Smart Achievers · Accepted Answer

(4)

$(\binom{n}{r}) + 2 (\binom{n}{r - 1}) + (\binom{n}{r - 2})$

$= [(\binom{n}{r}) + (\binom{n}{r - 1})] + [(\binom{n}{r - 1}) + (\binom{n}{r - 2})]$

$[Here [\begin{matrix} n \\ r \end{matrix}], [\begin{matrix} n \\ r - 1 \end{matrix}] and [\begin{matrix} n \\ r - 2 \end{matrix}] represent {}^{n}C_{r}, {}^{n}C_{r - 1} and {}^{n}C_{r - 2}]$

$= (\binom{n + 1}{r}) + (\binom{n + 1}{r - 1})$ $= (\binom{n + 2}{r})$ $[∵ {}^{n}{C_{r}} + {}^{n}{C_{r - 1}} = {}^{n + 1}{C_{r}}]$

Solution

Q.
For $2 \leq r \leq n, (\binom{n}{r}) + 2 (\binom{n}{r - 1}) + (\binom{n}{r - 2}) =$ [2000]

1 $(\binom{n + 1}{r - 1})$

2 $2 (\binom{n + 1}{r + 1})$

3 $2 (\binom{n + 2}{r})$

4 $(\binom{n + 2}{r})$

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Solution

Q. For 2≤r≤n,(nr)+2(nr-1)+(nr-2)= [2000]

1 (n+1r-1)

2 2(n+1r+1)

3 2(n+2r)

4 (n+2r)