Question

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

Answer 1

Answer: 5050

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