Question

Let $a_1, a_2, \ldots \ldots ., a_n$ be real numbers such that $\sqrt{a_1}+\sqrt{a_2-1}+\sqrt{a_3-2}+\ldots \ldots+\sqrt{a_n-(n-1)}=\frac{1}{2}\left(a_1+a_2+\ldots \ldots+a_n\right)-\frac{n(n-3)}{4} .$ Compute the value of $\displaystyle\sum_{ i =1}^{100} a _{ i }$.

Answer 1

Let $\sqrt{a_1}=b_1 ; \sqrt{a_2-1}=b_2 ; \sqrt{a_3-2}=b_3$;.......
$\sqrt{a_n-(n-1)}=b_n$
$\therefore b_1+b_2+\ldots \ldots+b_n=\frac{1}{2}\left[b_1^2+\left(b_2^2+1\right)+\left(b_3^2+2\right)+\ldots \ldots+\left(b_n^2+(n-1)\right)\right]-\frac{n(n-3)}{4} $
$\therefore \sum b _{ i }=\frac{1}{2}\left[\left( b _1^2+ b _2^2+ b _3^2+\ldots+ b _{ n }^2\right)+(1+2+3+\ldots \ldots+( n -1)]-\frac{ n ( n -3)}{4}\right.$
$\Rightarrow 2 \sum b_i=\sum b_i{ }^2+\frac{n(n-1)}{2}-\frac{n(n-3)}{2} $
$\Rightarrow 2 \sum b_i=\sum b_i{ }^2+\frac{n(n-1)-n(n-3)}{2} $
$\Rightarrow 2 \sum b_i=\sum b_i{ }^2+n$
$\therefore \sum b _{ i }{ }^2-2 \sum b _{ i }+\sum 1=0 \Rightarrow \displaystyle\sum_{ i =1}^{ n }\left( b _{ i }-1\right)^2=0 \Rightarrow b _{ i }-1=0, \forall i =1,2, \ldots, n$
Putting $i =1,2,3, \ldots \ldots ., 100$
$b_1-1=0 \Rightarrow b_1^2=a_1=1 $
$b _2-1=0 \Rightarrow b _2^2= a _2-1=1 \Rightarrow a _2=2$
$b_3-1=0 \Rightarrow b_3^2=a_3-2=1 \Rightarrow a_3=3 \text { and so on }$
Hence $a_n=n$
$\therefore \displaystyle\sum_{ i =1}^{100} a _{ i }=1+2+3+\ldots \ldots .+100=5050$