# Thread: mathematical induction problem no.2

1. ## mathematical induction problem no.2

mathematical induction problem no.2
sorry for my laziness of not typing it!

2. 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

3. Hello, afeasfaerw23231233!

Here's another approach . . .

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

4. thanks. i shouldn't make it as [1x4-2]^3+[2x4-2]^3+...