Need derivations of some formulas - 2! Help!

  1. (n0)+(n1)+(n2)+.....+(nn)=2n\left( \begin{matrix} n \\ 0 \end{matrix} \right) +\left( \begin{matrix} n \\ 1 \end{matrix} \right) +\left( \begin{matrix} n \\ 2 \end{matrix} \right) +.....+\left( \begin{matrix} n \\ n \end{matrix} \right) \quad =\quad { 2 }^{ n }

  2. (n0)+(n+11)+(n+22)+.....+(n+rr)=(n+r+1r)\left( \begin{matrix} n \\ 0 \end{matrix} \right) +\left( \begin{matrix} n+1 \\ 1 \end{matrix} \right) +\left( \begin{matrix} n+2 \\ 2 \end{matrix} \right) +.....+\left( \begin{matrix} n+r \\ r \end{matrix} \right) \quad =\quad \left( \begin{matrix} n+r+1 \\ r \end{matrix} \right)

  3. (rr)+(r+1r)+(r+2r)+.....+(nr)=(n+1r+1)\left( \begin{matrix} r \\ r \end{matrix} \right) +\left( \begin{matrix} r+1 \\ r \end{matrix} \right) +\left( \begin{matrix} r+2 \\ r \end{matrix} \right) +.....+\left( \begin{matrix} n \\ r \end{matrix} \right) \quad =\quad \left( \begin{matrix} n+1 \\ r+1 \end{matrix} \right)

  4. Integrating factor - For dydx+Py=Q\frac{dy}{dx} + Py = Q, IF=ePdxIF = {e}^{\int { P } dx}

Extra Credit

If you can give me some tips for such combinatorical sums.

You can add more from your own side.

Note by Kartik Sharma
5 years ago

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1 vote

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here (x, y) = (1 , 1)

thus it is (1+1)n=2n( 1 + 1)^{n} = 2^{n}

  1. (n+r+1r)=(n+rr)+(n+rr1) \binom{n + r + 1}{r} = \binom{n + r}{r} + \binom{n + r}{r - 1}

(n+rr1)=(n+r1r1)+(n+r1r2) \binom{n + r }{r - 1} = \binom{n + r - 1}{r - 1} + \binom{n + r - 1}{r - 2}

(n+r1r2)=(n+r2r2)+(n+r3r3)\binom{n + r - 1}{r - 2} = \binom{n + r - 2}{r - 2 } + \binom{n + r - 3}{r - 3}

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(n+32)=(n+22)+(n+21)\binom{n+3}{2} = \binom{n + 2}{2} + \binom{n + 2}{1}

(n+21)=(n+11)+(n+10)\binom{n+2}{1} = \binom{n + 1}{1} + \binom{n + 1}{0} =(n+11)+(n0)= \binom{n + 1}{1} + \binom{n}{0}

Adding, we get,

(n0)+(n+11)+.............+(n+rr)=(n+r+1r)\binom{n}{0} + \binom{n + 1}{1} + ............. + \binom{n + r}{r} = \binom{n + r + 1}{r}

I think now you can derive the third @Kartik Sharma

Sorry i don't know the fourth

U Z - 5 years ago

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Yeah, I thought of the same! Try the others!

Kartik Sharma - 5 years ago

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Use fundamental formulas to derive these kinds of sums @Kartik Sharma

U Z - 5 years ago

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For - 1) ( Although Megh aleready post it But here one another approach for this )

Let P = Total Number's of ways of selecting n things (Say shoes ) in all possible ways

So Obviously , LHS = P

and also there is two possibility for one shoe for selection , i.e either it will be selected or not .

So all possible ways is " P " P=2×2×2×2......2=2nP\quad =\quad 2\times 2\times 2\times 2\quad .\quad .\quad .\quad .\quad .\quad .\quad 2\quad =\quad { 2 }^{ n }\quad

Deepanshu Gupta - 5 years ago

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Another method for 3. ( but i think the usual is better)

Let any r of n + r + 1 objects given. let it be a1,a2............,ar a_{1} , a_{2} ............ , a_{r}

Case 1 - it does not contain a1a_{1} , it will happen in (n+rr)\binom{n + r}{r} ways for the r things to choosen from the remaining n + r things.

Case 2 - it contains a1a_{1} but does not contain a2a_{2} . This will happen in (n+rr1)\binom{n + r - }{ r - 1} ways (we have chossen a1a_{1} and not a2a_{2} , we have only n + r - 1 things to choose from and we need only r - 1)

Case 3 - it contains a1,a2 a_{1} , a_{2} and not a3 a_{3} , can happen in (n+r2r2)\binom{n + r - 2}{r - 2} ways(same reasoning)

Case 4 , case 5 ...... etc

Case r - it contains a1,a2,......ar1a_{1} , a_{2} , ...... a_{r - 1} but not ara_{r} , can happen in (n+11)\binom{n + 1}{1} ways.

case r +1 - it contains all things , can happen in (n0)\binom{n}{0}ways .

Thus-

(n0)+(n+11)+.............+(n+rr)=(n+r+1r)\binom{n}{0} + \binom{n + 1}{1} + ............. + \binom{n + r}{r} = \binom{n + r + 1}{r}

U Z - 5 years ago

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