# Smart Algebraic Technique

Calculate:
\$latex (3+1). (3^2 +1). (3^4 + 1)(3^8 +1)…. (3^{32} +1)
\$

Let
\$latex x = (3+1). (3^2 +1). (3^4 + 1)(3^8 +1)…. (3^{2n} +1)
\$

Or:
\$latex displaystyle
x = sum_{n=0}^{n}(3^{2n} + 1) \$

Quite messy to expand out:

\$latex displaystyle {
sum_{n=0}^{n} (3^{2n})
+
sum_{n=0}^{n}(1)
= ….
}
\$

This 14-year-old vienamese student in Berlin – Hyyen Nguyen Thi Minh discovered a smart trick using the identity:
\$latex displaystyle { (a -1).(a + 1) = a^{2} – 1}\$
or more general,
\$latex displaystyle boxed {
(a^{n} -1).(a^{n} + 1) = a^{2n} – 1
}\$

He multiplies x by (3-1):

\$latex
x. (3-1) = (3-1)(3+1). (3^{2} +1)… (3^{2n} + 1)
\$
\$latex 2x = (3^{2} -1). (3^{2} +1)…(3^{2n} + 1)
\$

\$latex 2x = (3^{4} -1).(3^{4} +1) … (3^{2n} + 1)
\$
.
.
.

\$latex 2x = (3^{4n} -1) \$

\$latex displaystyle boxed
{
x = (3^{4n}…

View original post 25 more words

## Author: tomcircle

Math amateur

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