# series

• Sep 15th 2012, 10:23 AM
geniusgarvil
series
Prove that each term of the following sequence is a perfect square:
49 , 4489 , 444889 , 44448889............................
• Sep 15th 2012, 10:44 AM
HallsofIvy
Re: series
Your very first number is NOT a perfect square! You obviously mean 49, not 44. And then each succeeding number is gained by putting a new "4" and "8" into the center. Have you looked at the square root of each of those numbers? That should give you a hint.
• Sep 15th 2012, 12:02 PM
Soroban
Re: series
Hello, geniusgarvil!

Quote:

Prove that each term of the following sequence is a perfect square:

. . $49,\,4489,\,444889,\,44448889\,\hdots$

$\text{The }4^{th}\text{ term is: }\:a_4 \;=\;4444(10^4) + 888(10) + 9$

. . . . . . . . . . . . $=\;4(1111)(10^4) + 8(111)(10) + 9$

. . . . . . . . $=\;4\left(\frac{10^4-1}{9}\right)(10^4) + 8\left(\frac{10^3-1}{9}\right)(10) + 9$

$\text{The }n^{th}\text{ term is: }\:a_n \;=\;4\left(\frac{10^n-1}{9}\right)10^n + 8\left(\frac{10^{n-1}-1}{9}\right)10 + 9$

$\text{We have: }\:\tfrac{4}{9}(10^{2n}) - \tfrac{4}{9}(10^n) + \tfrac{8}{9}(10^n) - \tfrac{8}{9}(10) + 9$

. . . . . $=\;\tfrac{4}{9}(10^{2n}) + \tfrac{4}{9}(10^n) + \tfrac{1}{9}$

. . . . . $=\;\tfrac{1}{9}\big[4(10^{2n}) + 4(10^n) + 1\big]$

. . . . . $=\;\tfrac{1}{9}\big[2(10^n) + 1\big]^2$

. . . . . $=\;\left[\frac{2(10^n) + 1}{3}\right]^2 \;\;\hdots\;\;\text{ a square!}$
• Sep 15th 2012, 09:18 PM
geniusgarvil
Re: series
wow, that's really amazing. Thanx