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

  • 1
    $\begingroup$ Looks to me like the textbook is in error. $\sum_{n=1}^{100}n=5050$ may have something to do with the mistake. $\endgroup$
    – saulspatz
    Jul 24, 2018 at 16:35
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    $\begingroup$ You can type the solution in your question. $\endgroup$
    – saulspatz
    Jul 24, 2018 at 16:42
  • 2
    $\begingroup$ No. You should type your question and not post images. If it's not worth your time and trouble to type the question, why is it worth my time and trouble to type an answer? $\endgroup$
    – saulspatz
    Jul 24, 2018 at 16:54
  • 2
    $\begingroup$ Read my last comment. Stop sending me messages. I don't want to chat with you. $\endgroup$
    – saulspatz
    Jul 24, 2018 at 17:05
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    $\begingroup$ For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have included some text from the pictures into your post. Please, edit it further if needed. $\endgroup$ Jul 25, 2018 at 8:42

2 Answers 2


You are indeed given that


and upon substituting $n=100$, we get


It is correct.


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 $\sum_{i=1}^{100}a_i$.

  • $\begingroup$ by this method answer must be 4850 but answer in my textbook is 5050 $\endgroup$
    – Hercules
    Jul 24, 2018 at 16:33
  • $\begingroup$ which book are you using? regardless of which method, all correct methods should lead to the same solution. $\endgroup$ Jul 24, 2018 at 16:34
  • $\begingroup$ sir i can send u snapshot of its solution please clarify my doubt $\endgroup$
    – Hercules
    Jul 24, 2018 at 16:36
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    $\begingroup$ most likely it is an error in the book solution. there are mistakes in books. $\endgroup$ Jul 24, 2018 at 16:37
  • $\begingroup$ sir tell me anyplace where i can send u snapshot of solution given in my book $\endgroup$
    – Hercules
    Jul 24, 2018 at 16:38

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, $


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