Question

Let $P=\sum _{r=1}^{50}\frac{{}^{50+r}{C}_r(2r-1)}{{}^{50}{C}_r(50+r)},Q=\sum _{r=0}^{50}{(}^{50}{C}_r{)}^2$ and $R=\sum _{r=0}^{100}(-1{)}^r{(}^{100}{C}_r{)}^2$. Then

• A. $Q=R$
• B. $P-Q=-1$
• C. $Q+R=2P+1$
• D. $P-R=1$

Answer 1

Answer: (A), (B)

$P-Q=-1$

Let
$T\left(r\right)=\frac{{}^{50+r}{C}_r(2r-1)}{{}^{50}{C}_r(50+r)}=\frac{{}^{50+r}{C}_r}{{}^{50}{C}_r}(1-\frac{51-r}{50+r})=\frac{{}^{50+r}{C}_r}{{}^{50}{C}_r}-\frac{{}^{50+r}{C}_r}{{}^{50}{C}_r}\left(\frac{51-r}{50+r}\right)$

Now,
$\frac{{}^{50+r}{C}_r}{{}^{50}{C}_r}\left(\frac{51-r}{50+r}\right)=\frac{(50+r){}^{50+r-1}{C}_{r-1}}{\left(r\right)}\times \frac{(50-r)!r!}{50!}\times \left(\frac{51-r}{50+r}\right)=\frac{{}^{50+r-1}{C}_{r-1}}{1}\times \frac{(51-r)!(r-1)!}{50!}=\frac{{}^{50+r-1}{C}_{r-1}}{{}^{50}{C}_{r-1}}$
So,
$T\left(r\right)=\frac{{}^{50+r}{C}_r}{{}^{50}{C}_r}-\frac{{}^{50+r-1}{C}_{r-1}}{{}^{50}{C}_{r-1}}\Rightarrow T\left(r\right)=V\left(r\right)-V(r-1)$,
where $V\left(r\right)=\frac{{}^{50+r}{C}_r}{{}^{50}{C}_r}$

Now sum of the given series,
$P=\sum _{r=1}^{50}T\left(r\right)$
$=V\left(50\right)-V\left(0\right)$
$={}^{100}{C}_{50}-1$

$Q=\sum _{r=0}^{50}{(}^{50}{C}_r{)}^2$
We know that,
$(1+x{)}^{50}(x+1{)}^{50}=(1+x{)}^{100}\sum _{r=0}^{50}{}^{50}{C}_r{x}^r\times \sum _{r=0}^{50}{}^{50}{C}_r{x}^{n-r}=\sum _{r=0}^{100}{}^{100}{C}_r{x}^r$
Comparing the coefficient of $x^{50}$,
$\sum _{r=0}^{50}{\left({}^{50}{C}_r\right)}^2={}^{100}{C}_{50}$
Therefore,
$\Rightarrow Q={}^{100}{C}_{50}$

Now,
$R=\sum _{r=0}^{100}(-1{)}^r{(}^{100}{C}_r{)}^2$
We know that,
$(x+1{)}^{100}(1-x{)}^{100}=(1-x^2{)}^{100}\sum _{r=0}^{100}{}^{100}{C}_r{x}^{n-r}\times \sum _{r=0}^{100}(-1{)}^r{}^{100}{C}_r{x}^r=\sum _{r=0}^{100}(-1{)}^r{}^{100}{C}_r(x^2{)}^r$
Comparing the coefficient of $x^{100}$,
$\sum _{r=0}^{100}(-1{)}^r{(}^{100}{C}_r{)}^2=(-1{)}^{50}{}^{100}{C}_{50}$

$\therefore R={}^{100}{C}_{50}$

Therefore,
$P={}^{100}{C}_{50}-1Q={}^{100}{C}_{50}R={}^{100}{C}_{50}$