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Let $a,b$ are non-negative integers.

Let define

$$S(a)=1+2+3+...+a=\sum_{i=0}^{a} i$$

and $$S(b)=1+2+3+...+b=\sum_{i=0}^{b} i$$

Question

Show that each $k\in \mathbb{N}$ and $k \ne 2^t$ $ \forall t\in \mathbb{N} $ can be represent as

$$ k = S(a)-S(b)$$ where $a \ne b+1$

My attempted

We can write $k$ as

$$ k = S(a)-S(b)$$

$$= n+(n+1)+...+(n+u)= \sum_{i=0}^{u}n+i$$

$$=\frac{(u+1)(2n+u)}{2}$$ Where $u\ge 1$

For $k>1$ and $k$ is odd number then $k$ can be represent as

$k= 2r+1 = r+(r+1)= S(r+1)-S(r-1)$

Now proof for, if $k= 2^t$ then $ k\ne S(a)-S(b)$

Proof

Let suppose $$k = n+(n+1)+...+(n+u)$$

$$=\frac{(u+1)(2n+u)}{2}= 2^t$$

So $$ (u+1)(2n+u)= 2^{t+1}$$

Case1

if $u$ is odd then $u+1=$ even and $2n+u =$ odd so even×odd $\ne 2^{t+1}$ because $ 2^{t+1}$ content only even multiples except $1$.

Case2

if $u$ is even then $u+1=$ odd and $2n+u =$ even so odd×even $\ne 2^{t+1}$ similarly as case 1

So both cases show complete proof for $k \ne 2^t$

Example

$6=1+2+3, 10=1+2+3+4,$

$ 12=3+4+5, 14=2+3+4+5, $

$18=5+6+7, 20=2+3+4+5+6$

$22=6+7+8, 24=7+8+9,$

$ 26=5+6+7+8,...$

You can check advanced similar problem here

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  • $\begingroup$ it is ok.............. $\endgroup$ – Aqua Sep 22 at 7:38
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You have shown above that it is impossible for any number of the form $2^t$ to be expressed as $S(a)-S(b)$. You have also shown that it is possible for all odd numbers. All you have to do is to extend the same idea for even values.

Case 1: The number is of the form $2^{t-1}m$ where $m \geqslant 2^t+1$ is odd

$$\frac{2^t \cdot (2^t+1)}{2}=2^{t-1}(2^t+1)=1+2+\cdots+2^t$$ $$2^{t-1}({2^t+1}+2k)=(1+k)+(2+k)+\cdots+(2^t+k)=S(2^t+k)-S(k)$$ for all non-negative integers $k$. We set $k=m-2^t-1$

Case 2: The number is of the form $2^{t-1}m$ where $1<m \leqslant 2^t-1$ is odd $$\frac{m(m+1)}{2}=1+2+\cdots+m$$ $$m(\frac{m+1}{2}+k)=(1+k)+(2+k)+\cdots+(m+k)=S(m+k)-S(k)$$ for all non-negative integers $k$. We set $k=2^t-1-m$

The reason why $m \neq 1$ is because we must have $a \neq b+1$ but at $m=1$, we have $a=k+1$ and $b=k$. Thus, using the two cases, any even number which is not a power of $2$ can be expressed as required.

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