Let postive integers sequence $\{a_n\}$,if for any postive integers $m$ we have $$\sum_{i=1}^{a_m}a_{i}=(2m-1)^2$$ show that $$a_n=2n-1$$
It seem can use mathematical induction? Now only following $$\sum_{i=1}^{a_m} a_i-\sum_{i=1}^{a_{m-1}}a_i=a_{a_{m-1}+1}+a_{a_{m-1}+2}+\cdots+a_{a_m}$$Thanks in advance!
First choose $m=1$ so : $$a_1+\ldots+a_{a_1}=1$$ but the $a_i$'s are positive integers so it must be that $a_1=1$ .
The sequence in obviously one-to-one because if for some $i,j$ we have $a_i=a_j$ then the sums on the LHS are the same so $(2i-1)^2=(2j-1)^2$ which leads to $i=j$ .
The sequence is also strictly increasing because if $i>j$ then the sum on the $i$-LHS is bigger than the sum on the $j$-LHS so $a_i>a_j$ .As a corollary we get :
$$a_n \geq n$$ for every $n$ .
If $a_2=2$ then choose $m=2$ which leads to $a_1+a_2=9$ which is false .
Also if $a_2 \geq 4$ then putting $m=2$ we get :
$9 \geq a_1+a_2+a_3+a_4 \geq1+2+3+4=10$ ,false so $a_2=3$ .
Now assume that we established $a_i=2i-1$ for every $i$ between $1$ and $n$ .Denote $a_{n+1}=x$ .
Subtract the equations for $m=n$ and $m=n+1$ to get : $$a_{2n}+\ldots+a_x=(2n+1)^2-(2n-1)^2=8n$$
with $x \geq 2n$ .
Assume that $x\geq 2n+2$ .Using the strictly increasing condition we get :
$a_{2n} \geq a_n+n=3n-1$ , $a_{2n+1} \geq 3n$ , $a_{2n+2} \geq 3n+1$ so plugging back : $$8n \geq a_{2n}+a_{2n+1}+a_{2n+2} \geq 9n$$ clearly false .
So it must be that $x \leq 2n+1$ .
Now assume that $x=2n$ so $a_{2n}=8n$ .
Now use $m=2n$ and subtract it with $m=n$ : $$(a_1+\ldots+a_{8n})-(a_1+\ldots+a_{2n-1})=(4n-1)^2-(2n-1)^2=4n(3n-1)$$
$$a_{2n}+\ldots+a_{8n}=4n(3n-1)$$
Use again the strictly increasing property to get : $$a_{2n+d} \geq a_{2n}+d=8n+d$$ for every $d$ so plugging back we get :
$$4n(3n-1) \geq 8n+(8n+1)+\ldots+14n=11n(6n+1)$$ which is clearly false .
It follows that $a_{n+1}=x=2n+1$ and from the inductive hypothesis that $$a_n=2n-1$$ for every $n$ .
We have $\{a_n\}$ with $a_n\gt0$ and $$\sum_{i=1}^{a_m}a_{i}=(2m-1)^2$$ Hence $$\sum_{i=1}^{a_m}a_{i}=a_1+a_2+\cdot\cdot\cdot+a_{a_m}=(2m-1)^2$$ $$\sum_{i=1}^{a_{m+1}}a_{i}=a_1+a_2+\cdot\cdot\cdot+a_{a_{m+1}}=(2m+1)^2$$ Suppose $a_m=r, a_{m+1}=s$ with $r \ge s; m\ge1$ so that $a_1+a_2+\cdot\cdot\cdot+a_r=(2m-1)^2\ge a_1+a_2+\cdot\cdot\cdot+a_s=(2m+1)^2\Rightarrow -8m\ge0$ This can't be true because $m\gt0$. Hence the sequence $\{a_n\}$ is strictly increasing. Besides$$\sum_{i=1}^{a_{m+1}}a_{i}-\sum_{i=1}^{a_{m-1}}a_{i}=8m$$ $$\boxed {a_1=k_1=?}$$
$$\sum_{i=1}^{a_1}a_{i}=a_1+a_2+\cdot\cdot\cdot+a_{k_1}=(2\cdot1-1)^2=1\Rightarrow k_1=1\Rightarrow a_1=1$$ because $a_n\gt 0$. $$\sum_{i=1}^{a_2}a_{i}=a_1+a_2+a_3+\cdot\cdot\cdot+a_{a_2}=(2\cdot2-1)^2=9$$ where $1+2+3+4+5+\cdot\cdot\cdot +a_{a_2}\le a_1+a_2+a_3+\cdot\cdot\cdot+a_{a_2}$. Therefore $2\le {a_2}\le3$ because $1+2+3\lt9\lt1+2+3+4$. In order to have $\boxed {a_n=2n-1}$ only $a_2=3$ agrees in which case $a_1+a_2+a_3=9$ hence $(a_1,a_2,a_3)=(1,3,5)$ Now $$\sum_{i=1}^{a_{3}}a_{i}=a_1+a_2+a_3+a_4+a_5=(2\cdot3-1)^2=25$$ i.e.$$a_4+a_5=25-9=16$$ where $a_4\ge6$ so $(a_4,a_5)=(6,10),(7,9)$ and only $(7,9)$ agrees. Similarly $$\sum_{i=1}^{a_{4}}a_{i}=a_1+a_2+a_3+a_4+a_5+a_6+a_7=49$$ i.e.$$a_6+a_7=49-25=24$$ where $a_6\ge10\Rightarrow(a_6,a_7)=(10,14),(11,13)$ and $(a_6,a_7)=(11,13)$.
NOTATION: For easy, $\boxed {{\sum_{i=1}^{a_{n}}a_{i}=S(a_{n})}}$
I use these first seven calculated values of $a_n$ to infer an expression of S (a_n) as a function of n.
$$S(a_1)=a_1=1$$ $$S(a_2)=S(a_1)+a_2+a_3=9\Rightarrow a_2+a_3=8\cdot1$$
$$S(a_3)= S(a_2) +a_4+a_5=25\Rightarrow a_4+a_5=8\cdot2$$
$$S(a_4)=S(a_3)+a_6+a_7=49\Rightarrow a_6+a_7=8\cdot3$$
$$S(a_5)= S(a_4)+a_8+a_9=81\Rightarrow a_8+a_9=8\cdot4$$ $$S(a_6)=S(a_5)+a_{10}+a_{11}=121 \Rightarrow a_{10}+a_{11}=8\cdot5$$ $$S(a_7)=S(a_6)+a_{12}+a_{13}=169\Rightarrow a_{12}+a_{13}=8\cdot6$$
These seven equalities successively generated show at first sight that $a_1=1,a_2=3,…..,a_7=13$ because, according to the statement, the last index of the $\sum_{i=1}^{a_{n}}a_{i}$ must be $a_1,a_2,a_3,….a_7$ respectively. Each value of an $a_i$ determines the value of $a_{i+1}=a_i+2$ which then verify that $\boxed {a_n=2n-1}$.
On the other hand, we have by easy induction on $S(a_n)-S(a_{n-1})=8(n-1)$,
$$S(a_n)=1+8[1+2+3+\cdot\cdot\cdot+(n-1)]$$
i.e. $$S(a_n)=1+4n(n-1)=(2n-1)^2\iff \sum_{i=1}^{a_m}a_{i}=(2m-1)^2 $$
(BIS proof).-Optionally, using induction hypothesis $\sum_{i=1}^{a_m}a_{i}=(2m-1)^2$, having $\sum_{i=1}^{a_1}a_{i}=(2\cdot1-1)^2=1$, it follows
$S(a_{n+1})-S(a_n)=8n\Rightarrow S(a_{n+1})=S(a_n)+8n=(2n-1)^2+8n=(2n+1)^2$, thus
$a_{n+1}=2n+1\iff a_{n+1}=2(n+1)-1$.
NOTE.- Finally, it was very difficult to interpret the statement at the beginning; I also thought, as well as distinguished followers of Stack Exchange there must be some typo. What I find remarkable is the equivalence for natural integers $\sum_{i=1}^{a_{n}}a_{i}=(2n-1)^2\iff a_n=2n-1$ which everyone knows only one implication.
