# mathematical induction problem no.2

• Jan 12th 2008, 01:16 AM
afeasfaerw23231233
mathematical induction problem no.2
mathematical induction problem no.2
sorry for my laziness of not typing it!
• Jan 12th 2008, 08:44 AM
topsquark
Quote:

Originally Posted by afeasfaerw23231233
mathematical induction problem no.2
sorry for my laziness of not typing it!

I don't know if this is how you are to do it, but my first thought is that your series:
$\displaystyle 2^3 + 6^3 + 10^3 + ~...~ + 38^3$

$\displaystyle = 2^3(1^3 + 3^3 + 5^3 + ~...~ + 19^3)$

Now consider
$\displaystyle 1^3 + 2^3 + 3^3 + ~...~ + 19^3 = \frac{19^2(19^2 + 1)}{4}$

$\displaystyle (1^3 + 3^3 + 5^3 + ~...~ + 19^3) + (2^3 + 4^3 + 6^3 + ~...~ + 18^3) = \frac{19^2(19^2 + 1)}{4}$

$\displaystyle (1^3 + 3^3 + 5^3 + ~...~ + 19^3) + 2^3(1^3 + 2^3 + 3^3 + ~...~ + 9^3) = \frac{19^2(19^2 + 1)}{4}$

And the second sum is just a sum of cubes, so
$\displaystyle (1^3 + 3^3 + 5^3 + ~...~ + 19^3) + 2^3 \left ( \frac{9^2(9^2 + 1)}{4} \right ) = \frac{19^2(19^2 + 1)}{4}$

$\displaystyle 1^3 + 3^3 + 5^3 + ~...~ + 19^3 = \frac{19^2(19^2 + 1)}{4} - 2^3 \left ( \frac{9^2(9^2 + 1)}{4} \right )$

And finally:
$\displaystyle 2^3 + 6^3 + 10^3 + ~...~ + 38^3 = 2^3(1^3 + 3^3 + 5^3 + ~...~ + 19^3) =$$\displaystyle 2^3 \left [ \frac{19^2(19^2 + 1)}{4} - 2^3 \left ( \frac{9^2(9^2 + 1)}{4} \right ) \right ]$

-Dan
• Jan 12th 2008, 07:05 PM
Soroban
Hello, afeasfaerw23231233!

Here's another approach . . .

Quote:

Given: .$\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3\;=\;\frac{n^2(n+1)^2}{4}$ . for all positive integers $\displaystyle n$

find the value of: .$\displaystyle A \;= \;2^3 + 6^3 + 10^3 + \cdots + 38^3$

We have: .$\displaystyle S \;=\;(2\cdot1)^3 + (2\cdot3)^3 + (2\cdot5)^3 + \cdots + (2\cdot19)^3$

. . . . . . . . . $\displaystyle = \;(2^3\cdot1^3) + (2^3\cdot3^3) + 2^3\cdot5^3) + \cdots + (2^3\cdot19^3)$

. . . . . . . . . $\displaystyle = \;8(1^3 + 3^3 + 5^3 + \cdots + 19^3)$

Let $\displaystyle H \:=\:1^3 + 3^3 + 5^3 + \cdots + 19^3$

Let $\displaystyle J \:=\:1^3 + 2^3 + 3^3 + \cdots + 20^3\:=\:\frac{20^2\!\cdot\!21^2}{4} \:=\:44,100$
Let $\displaystyle K \:=\:2^3 + 4^3 + 6^3 + \cdots + 20^3 \:=\;2^3(1^3 + 2^3 + \cdots + 10^3) \:=\:8\cdot\frac{10^2\!\cdot\!11^2}{4} \:=\:24,200$

Hence: .$\displaystyle H \;=\;J - K \;=\;44,100 - 24,200 \:=\:19,900$

Therefore: .$\displaystyle S \;=\;8(19,900) \;=\;159,200$

• Jan 12th 2008, 07:45 PM
afeasfaerw23231233
thanks. i shouldn't make it as [1x4-2]^3+[2x4-2]^3+...