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Mathematics Question on Binomial theorem

If ∑R=0n(−1)r  nCrr+3Cr=3a+3\displaystyle\sum^n_{R = 0} (-1)^r \; \frac{^{n}C_{r}}{^{r+3}C_{r}} = \frac{3}{a+3} , then a - n is equal to

A

0

B

1

C

2

D

None of these

Answer

0

Explanation

Solution

nCrr+3Cr=3!1(r+3)(r+2).nCr(r+1)\frac{^{n}C_{r}}{^{r+3}C_{r}} = 3! \frac{1}{\left(r+3\right)\left(r+2\right)} . \frac{^{n}C_{r}}{\left(r+1\right)}
=3!1(r+3)(r+2).n+1Cr+1(n+1)= 3! \frac{1}{\left(r+3\right)\left(r+2\right)} . \frac{^{n+1}C_{r+1}}{\left(n+1\right)}
=3!1(r+3)(n+1)n+1Cr+1r+2= 3! \frac{1}{\left(r+3\right)\left(n+1\right)} \frac{^{n+1}C_{r+1}}{r+2}
=3!1(r+3)(n+1)n+2Cr+2n+2= 3! \frac{1}{\left(r+3\right)\left(n+1\right)} \frac{^{n+2}C_{r+2}}{n+2}
=3!(n+1)(n+2).n+2Cr+2r+3= \frac{3!}{\left(n+1\right)\left(n+2\right)} . \frac{^{n+2}C_{r+2}}{r+3}
=3!(n+1)(n+2)(n+3)n+3Cr+3= \frac{3!}{\left(n+1\right)\left(n+2\right)\left(n+3\right)} {^{n+3}C_{r+3} }
∴∑r=0n(−1)rnCrr+3Cr\therefore \displaystyle\sum^{n}_{r=0} \left(-1\right)^{r} \frac{^{n}C_{r}}{^{r+3}C_{r}}
=6(n+1)(n+2)(n+3)∑r=0n(−1)rn+3Cr+3= \frac{6}{\left(n+1\right)\left(n+2\right)\left(n+3\right)} \displaystyle\sum^{n}_{r=0} \left(-1\right)^{r} {^{n+3}C_{r+3}}
=6(n+1)(n+2)(n+3)[n+3C3−n+3C4+...+(−1)nn+3Cn+3]= \frac{6}{\left(n+1\right) \left(n+2\right)\left(n+3\right)} \left[^{n+3}C_{3} - {^{n+3}}C_{4} + ... + \left(-1\right)^{n } {^{n+3}}C_{n+3}\right]
=6(n+1)(n+2)(n+3)[n+3C0−n+3C1+n+3C2]= \frac{6}{\left(n+1\right) \left(n+2\right)\left(n+3\right)} \left[^{n+3}C_{0} - {^{n+3}C_{1} } + {^{n+3}C_{2}}\right]
[∵n+3C0−n+3C1+......+(−1)n+3×n+3Cn+3=0]\left[\because {^{n+3}C_{0} } - {^{n+3}C_{1}} + ...... +\left(-1\right)^{n+3} \times {^{n+3}C_{n+3}} = 0\right]
=6(n+1)(n+2)(n+3)(1−n−3+(n+3)(n+2)2)= \frac{6}{\left(n+1\right)\left(n+2\right)\left(n+3\right)} \left(1-n-3+ \frac{\left(n+3\right)\left(n+2\right)}{2}\right)
=3(n+1)(n+2)(n+3)(n2+3n+2)= \frac{3}{\left(n+1\right)\left(n+2\right)\left(n+3\right)} \left(n^{2} + 3n+2\right)
=3n+33n+3= \frac{3}{n+3} \frac{3}{n+3}
=3a+3⇒n= \frac{3}{a+3} \Rightarrow n
=a⇒a−n= a \Rightarrow a - n
=0= 0