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 $a_n=n$.
$$\therefore \sum_{i=1}^{100} a_i = 1+2+3+\dots+100=5050.$$
Best Answer
You are indeed given that
$$\sum_{i=1}^{n}a_i=\frac{n(n-3)}{2}$$
and upon substituting $n=100$, we get
$$\frac{100(97)}{2}=4850$$
It is correct.
Remark:
The correction is the second equality should be a minus sign, from there we can prove that $a_n=n$ and hence the sum of the first $100$ positive integers is $5050$.
That is the actual question is