If $^{n} C_{n - r} + 3^{n} C_{n - r + 1} + 3^{n} C_{n - r + 2} +^{n} C_{n - r + 3} =^{p} C_{r}$ then prove that $p = n + 3$ .

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Solution

$(^{n} C_{n - r} +^{n} C_{n - r + 1}) + 2 (^{n} C_{n - r + 1} +^{n} C_{n - r + 2}) + (^{n} C_{n - r + 2} +^{n} C_{n - r + 3})$
= $^{n + 1} C_{n - r + 1} + 2^{n + 1} C_{n - r + 2} +^{n + 1} C_{n - r + 3}$
= $^{n + 1} C_{n - r + 1} +^{n + 1} C_{n - r + 2} +^{n + 1} C_{n - r + 2} +^{n + 1} C_{n - r + 3}$
= $^{n + 2} C_{n - r + 2} +^{n + 2} C_{n - r + 3}$
= $^{n + 3} C_{n - r + 3}^{n + 3} C_{r}$ [ $∴^{n} C_{r} =^{n} C_{n - r}$ ]
$∴ p = n + 3.$

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If nCn−r+3nCn−r+1+3nCn−r+2+nCn−r+3=pCr then prove that p=n+3.

(nCn−r+nCn−r+1)+2(nCn−r+1+nCn−r+2)+(nCn−r+2+nCn−r+3)= n+1Cn−r+1+2n+1Cn−r+2+n+1Cn−r+3= n+1Cn−r+1+n+1Cn−r+2+n+1Cn−r+2+n+1Cn−r+3= n+2Cn−r+2+n+2Cn−r+3= n+3Cn+3n−r+3Cr [∴nCr=nCn−r]∴p=n+3.

If $^{n} C_{n - r} + 3^{n} C_{n - r + 1} + 3^{n} C_{n - r + 2} +^{n} C_{n - r + 3} =^{p} C_{r}$ then prove that $p = n + 3$ .