Let $a_1,a_2,\dots,a_n$ be real numbers such that $$\sqrt{a_1}+\sqrt{a_2-1}+\sqrt{a_3-2}+\dots+\sqrt{a_n-(n-1)}=\frac12(a_1+a_2+\dots+a_n)=\frac{n(n-3)}4$$ Compute the value of $\sum_{i=1}^{100} a_i$.
Can't I just use simply equation (i) to get $\sum_{i=1}^{100}a_i = \frac{(n)(n-3)}{2}$ and by putting $n=100$ I get its summation = $4850$
Why is it wrong ? please anybody explain me ?
The solution given in the book is as follows:
Let $\sqrt{a_1}=b_1$ \begin{align*} \sqrt{a_2-1}&=b_2 \\ \sqrt{a_3-2}&=b_3 \\ \ldots\\ \sqrt{a_n-(n-1)}&=b_n \\ \end{align*} \begin{align*} \therefore & b_1 + b_2 + \dots + b_n = \\ & \frac12 \left[b_1^2+(b_2^2+1)+\dots+(b_n^2-(n-1))\right] - \frac{n(n-3)}4 \end{align*} \begin{align*} \therefore & \sum b_i = \frac12 [(b_1^2+b_2^2+\dots+b_n^2)]+ \\ & (1+2+3+\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)}4 \\ \Rightarrow & 2\sum b_i = \sum b_i^2 + n \\ \Rightarrow & \sum b_i^2 - 2\sum b_i + \sum 1 = 0 \\ & b_1-1 =0 \qquad\Rightarrow\qquad b_1^2=a_1=1 \\ & b_2-1=0 \qquad\Rightarrow\qquad b_2^2 = a_2-1 = 1 \qquad\Rightarrow\qquad a_2=2\\ & b_3-1=0 \qquad\Rightarrow\qquad b_3^2 = a_3-2 = 1 \qquad\Rightarrow\qquad a_3=3 \end{align*} and so on. Hence $a_n=n$. $$\therefore \sum_{i=1}^{100} a_i = 1+2+3+\dots+100=5050.$$