Question

where $C_r$ represents coefficient of $x^r$ in the binomial expansion of $(1+x{)}^{100}$. If $S_1+S_2+S_3=a^b$ where $a,b\in N$ then find the least value of $(a+b)$.

Answer 1

Answer: 66

We have $S_1+S_2+S_3=\sum _{i=0}^{100}\sum _{j=0}^{100}{C}_i{C}_j={\left({}^{100}{C}_0+{}^{100}{C}_1+{}^{100}{C}_2+\dots \dots .+{}^{100}{C}_{100}\right)}^2=$${\left(2^{100}\right)}^2=2^{200}$
where $S_1=S_2=\frac{1}{2}\underset{0\le i\ne j\le 100}{\sum \sum }{C}_i{C}_j=\frac{1}{2}\left(2^{200}-S_3\right)$
[Revision test-1, 11th 2009-2010]
and
$S_3=\underset{0\le i=j\le 100}{\sum \sum }{C}_i{C}_j=\sum _{i=0}^{100}{C}_i^2={}^{100}{C}_0^2+{}^{100}{C}_1^2+\dots ..+{}^{100}{C}_{100}^2={}^{200}{C}_{100}$
(Using : ${}^n{C}_0{}^2+{}^n{C}_1{}^2+{}^n{C}_2{}^2+\dots \dots ..+{}^n{C}_n{}^2={}^{2n}{C}_n$ )
Now $S_1+S_2+S_3=2^{200}=4^{100}=16^{50}=256^{25}=\dots =a^b$
Hence least value of $(a+b)=16+50=66$