Maths Problem (Medium): Proving the Solution

[alice22]

Use English to explain what this is showing.
Also prove it!

 

[alice22]

Use English to explain what this is showing.
Also prove it!

I found this else where, so some of you may have seen it before.

 

[Klockan3]

Just expand the square and you are basically done. The i's in the rhs are the number of combinations which can produce that exact exponent.

 

[HallsofIvy]

Use English to explain what this is showing.
Also prove it!

The simplest thing to do is to look at what it says for some small n- say n= 3.
$$\left(\sum_{i=0}^{3-1}10^i\right)^2= \left(10^0+ 10^1+ 10^2\right)^2= (1+ 10+ 100)^2= 111^2$$.

$$\sum_{i=1}^n i10^{i-1}= 1(10^0)+ 2(10^1)+ 3(10^2)= 321$$

$$\sum_{i=1}^{n-1} i10^{2n-i-1}= 1(10^4}+ 2(10^3)= 1200$$

It's easy to calculate that $111^2= 12321= 1200+ 321$.

$11^2= 121$, $111^3= 12321$, $1111^2= 1234321$ , etc.

Do you see the pattern?
$$\sum_{i=1}^{n-1} i10^{2n-i-1}$$
is the first part- the 1234... Do you see how it is counting "down" because of the $10^{2n-i-1}$?
$$\sum_{i=1}^n i10^{i-1}$$
is the last part: 321

 

[alice22]

The simplest thing to do is to look at what it says for some small n- say n= 3.
$$\left(\sum_{i=0}^{3-1}10^i\right)^2= \left(10^0+ 10^1+ 10^2\right)^2= (1+ 10+ 100)^2= 111^2$$.

$$\sum_{i=1}^n i10^{i-1}= 1(10^0)+ 2(10^1)+ 3(10^2)= 321$$

$$\sum_{i=1}^{n-1} i10^{2n-i-1}= 1(10^4}+ 2(10^3)= 1200$$

It's easy to calculate that $111^2= 12321= 1200+ 321$.

$11^2= 121$, $111^3= 12321$, $1111^2= 1234321$ , etc.

Do you see the pattern?
$$\sum_{i=1}^{n-1} i10^{2n-i-1}$$
is the first part- the 1234... Do you see how it is counting "down" because of the $10^{2n-i-1}$?
$$\sum_{i=1}^n i10^{i-1}$$
is the last part: 321

Yes that's basically it the first bit gives the last n digits and the first bit gives the first n+1 digits where the length is 2n+1