If nCr+3 nCr+1+3 nCr+2+ nCr+3 nCr+4 nCr+1+6 nCr+2+4 nCr+3+ nCr+4=r+kn+k, then find the value of k

[(nCr+nCr+1)+2(nCr+1+nCr+2)+(nCr+2+nCr+3)][(nCr+nCr+1)+3(nCr+1+2nCr+2+nCr+3)+(nCr+3+nCr+4)]= n+1Cr+1+2(n+1Cr+2)+n+1Cr+3 n+1Cr+1+3(n+1Cr+2)+n+1Cr+3+n+1Cr+4=(n+1Cr+1+n+1Cr+2)+(n+1Cr+2+n+1Cr+3)(n+2Cr+2+n+2Cr+2)+2(n+2Cr+3)+n+2Cr+4= n+2Cr+2+n+2Cr+3(n+2Cr+2+n+2Cr+3)+(n+2Cr+3)+n+2Cr+4= n+3Cr+3 n+4Cr+4= n+3Cr+3 n+3Cr+3(r+4)n+4=r+4n+4∴r+4n+4=r+kn+k⇒k=4

If nCr+3 nCr+1+3 nCr+2+ nCr+3 nCr+4 nCr+1+6 nCr+2+4 nCr+3+ nCr+4=r+kn+k, then find the value of k

[(nCr+nCr+1)+2(nCr+1+nCr+2)+(nCr+2+nCr+3)][(nCr+nCr+1)+3(nCr+1+2nCr+2+nCr+3)+(nCr+3+nCr+4)]= n+1Cr+1+2(n+1Cr+2)+n+1Cr+3 n+1Cr+1+3(n+1Cr+2)+n+1Cr+3+n+1Cr+4=(n+1Cr+1+n+1Cr+2)+(n+1Cr+2+n+1Cr+3)(n+2Cr+2+n+2Cr+2)+2(n+2Cr+3)+n+2Cr+4= n+2Cr+2+n+2Cr+3(n+2Cr+2+n+2Cr+3)+(n+2Cr+3)+n+2Cr+4= n+3Cr+3 n+4Cr+4= n+3Cr+3 n+3Cr+3(r+4)n+4=r+4n+4∴r+4n+4=r+kn+k⇒k=4

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If $\frac{^{n} C_{r} + 3 (^{n} C_{r + 1}) + 3 (^{n} C_{r + 2}) +^{n} C_{r + 3}}{^{n} C_{r} + 4 (^{n} C_{r + 1}) + 6 (^{n} C_{r + 2}) + 4 (^{n} C_{r + 3}) +^{n} C_{r + 4}}$ $= \frac{r + k}{n + k},$ then find the value of k

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answer is D.

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Detailed Solution

$\frac{[(^{n} C_{r} +^{n} C_{r + 1}) + 2 (^{n} C_{r + 1} +^{n} C_{r + 2}) + (^{n} C_{r + 2} +^{n} C_{r + 3})]}{[(^{n} C_{r} +^{n} C_{r + 1}) + 3 (^{n} C_{r + 1} + 2^{n} C_{r + 2} +^{n} C_{r + 3}) + (^{n} C_{r + 3} +^{n} C_{r + 4})]}$

$= \frac{^{n + 1} C_{r + 1} + 2 (^{n + 1} C_{r + 2}) +^{n + 1} C_{r + 3}}{^{n + 1} C_{r + 1} + 3 (^{n + 1} C_{r + 2}) +^{n + 1} C_{r + 3} +^{n + 1} C_{r + 4}}$

$= \frac{(^{n + 1} C_{r + 1} +^{n + 1} C_{r + 2}) + (^{n + 1} C_{r + 2} +^{n + 1} C_{r + 3})}{(^{n + 2} C_{r + 2} +^{n + 2} C_{r + 2}) + 2 (^{n + 2} C_{r + 3}) +^{n + 2} C_{r + 4}}$