4+44+444...n terms = a, then find 81a/4+10+9n=? Please explain the concept behind this.

4 + 44 + 444 + 4444 + .... + 4444... (n terms)

This can be written as: 4*(1 + 11 + 111 + ....)
= 4*( 1 + 10 + 1 + 100 + 10 + 1 + ....)

= 4*(1 * 10$$^{n-1}$$ + 2 * 10$$^{n-2}$$ + 3 * 10$$^{n-3}$$ +....+ n * 10$$^{n-n}$$)

= 4*( (10$$^0$$ + ... + 10^$${n-1}$$) + (10$$^0$$ + ... + 10$$^{n-2}$$) + ... + (10$$^0$$))

= 4*((10$$^n$$ - 1)/9 + (10$$^{n-1}$$ - 1)/9 + .... + (10 - 1)/9)

= 4*(1/9) * 10*(10$$^n$$ - 1)/9 - n*(4/9) => a = (4/81)*10*(10$$^n$$ - 1) - 4n/9

So, 81a/4 = 10*(10$$^n$$ - 1) - 9n = 10$$^{n+1}$$ - 10 - 9n

81a/4 + 10 + 9n = 10$$^{n+1}$$