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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 $(a+b)$.

Binomial Theorem

Solution:

We have $S _1+ S _2+ S _3=\displaystyle\sum_{ i =0}^{100} \displaystyle\sum_{ j =0}^{100} C _{ i } C _{ j }=\left({ }^{100} C _0+{ }^{100} C _1+{ }^{100} C _2+\ldots \ldots .+{ }^{100} C _{100}\right)^2=$$\left(2^{100}\right)^2=2^{200}$
where $S _1= S _2=\frac{1}{2} \underset{0 \leq i \neq j \leq 100}{\sum\sum }C _{ i } C _{ j }=\frac{1}{2}\left(2^{200}- S _3\right)$
[Revision test-1, 11th 2009-2010]
and
$S _3=\underset{0 \leq i = j \leq 100} {\sum \sum} C _{ i } C _{ j }=\displaystyle\sum_{ i =0}^{100} C _{ i }^2={ }^{100} C _0^2+{ }^{100} C _1^2+\ldots . .+{ }^{100} C _{100}^2={ }^{200} C _{100}$
(Using : ${ }^{ n } C _0{ }^2+{ }^{ n } C _1{ }^2+{ }^{ n } C _2{ }^2+\ldots \ldots . .+{ }^{ n } C _{ n }{ }^2={ }^{2 n } C _{ n }$ )
Now $S _1+ S _2+ S _3=2^{200}=4^{100}=16^{50}=256^{25}=\ldots= a ^{ b }$
Hence least value of $(a+b)=16+50=66$