# 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:
$2^3 + 6^3 + 10^3 + ~...~ + 38^3$

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

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

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

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

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

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

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

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

Here's another approach . . .

Quote:

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

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

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

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

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

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

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

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

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

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