Finding $\sum_{i=1}^{100} a_i$ given that $\sqrt{a_1}+\sqrt{a_2-1}+\sqrt{a_3-2}+\dots+\sqrt{a_n-(n-1)}=\frac12(a_1+a_2+\dots+a_n)=\frac{n(n-3)}4$

Question

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 $a_n=n$. $$\therefore \sum_{i=1}^{100} a_i = 1+2+3+\dots+100=5050.$$

Answer 1

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

Let $a_1, a_2, \ldots , a_n$ be real numbers such that \begin{align}&\sqrt{a_1}+\sqrt{a_2-1}+\sqrt{a_3-2}+\ldots + \sqrt{a_n-(n-1)} =\\ &\frac12 (a_1+a_2+\ldots+a_n)\color{red}-\frac{n(n-3)}4\end{align} Compute the value of $\sum_{i=1}^{100}a_i$.

Answer 2

Another approach is to use the am-gm inequality As Siong Thye Goh has already mentioned, there is a typo in the exercise.

The i´th summand of the LHS is $\sqrt{[a_i-(i-1)]\cdot 1}$

Applying am-gm we get $\sqrt{[a_i-(i-1)]\cdot 1}\leq \frac{a_i-i+1+1}{2}=\frac{a_i-i+2}{2}$

Now we can sum up the terms:

$$\sum_{i=1}^n \frac{a_i-i+2}{2}=-\frac{n(n-3)}{2}+\sum_{i=1}^n \frac{a_i}{2}$$

Now we have the following statement:

The equality holds if $b_1=b_2=\ldots =b_n\quad \forall b_i \in \mathbb R^+$

Consequently the equality is true if $a_n=n, a_{n-1}=n-1, \ldots, a_1=1, $