1. ## Hard Series Question

Hi

For this question:

The sequences {1, 3, 5, ……,
p} and {1, 3, 5, ………, q} contain the integer values of

p
and q respectively.

Find the value of
p+q if :
{1 + 3 + 5 + ……+ p} +{1 + 3 + 5 + …… + q} = {1 + 3 + 5 + ……+ 33}

I've gotten this far:

0.5 (p+1) [(p+1)/2] + 0.5 (q+1) [(q+1)/2] = 0.5 (33+1) [(33+1)/2]

...

(p+1)^2 + (q+1)^2 = 34^2

What do I do now to get p + q???

Thank you!

2. You are correct that $(p + 1)^2 + (q + 1)^2 = 34^2$.

However, there is not any Pythagorean Triad which satisfies this equation. So I would say that this is unsolvable.

3. Originally Posted by kashuv
hi... i just joine this site... could u guide me as to how to start a thread? would be very greatful...thanks

4. $16^2+30^2=34^2$

so p=15 q=29

5. Originally Posted by UltraGirl
Hi

For this question:

The sequences {1, 3, 5, ……,
p} and {1, 3, 5, ………, q} contain the integer values of

p
and q respectively.

Find the value of
p+q if :
{1 + 3 + 5 + ……+ p} +{1 + 3 + 5 + …… + q} = {1 + 3 + 5 + ……+ 33}

I've gotten this far:

0.5 (p+1) [(p+1)/2] + 0.5 (q+1) [(q+1)/2] = 0.5 (33+1) [(33+1)/2]

...

(p+1)^2 + (q+1)^2 = 34^2

What do I do now to get p + q???

Thank you!

You could try...

sum of an arithmetic series is......... $average(number\ of\ terms)$

$1+3+5+....+p=\left(\frac{p+1}{2}\right)^2$

$1+3+5+....+q=\left(\frac{q+1}{2}\right)^2$

$1+3+5+....+33=17+17+17+...+17=17^2$

Now we can ask if $17^2$ is the sum of 2 squares.

You can do this by drawing a 17 by 17 square.
Then if we isolate a 16 by 16 square inside it, we are left with 1(17)+1(16)=33

If we isolate a 15 by 15 square, we have 2(17)+2(15)=34+30=64

That's a square, so we have $17^2=15^2+8^2$

Hence p (or q) is 29... from $\frac{p+1}{2}=15$

and the other (in this case q) is $\frac{q+1}{2}=8\ \Rightarrow\ q=15$

Hence p+q=44 is a solution, for p=29 and q=15.
However, that will not hold for other values of p+q that give 44 obviously,
since clearly p or q cannot be >33 among other things.

you can try a few more combinations to see if there are other solutions

$17^2=14^2+3(17)+3(14)$

$17^2=13^2+4(17)+4(13)$

$17^2=12^2+5(17)+5(12)$ etc

crikey!! Krahl got there first, was i really that long typing this??

6. Hello, UltraGirl!

Find the value of $p+q$ if

$(1\!+\!3\!+\!5\!+ \hdots +\!p) + (1\!+\!3\!+\!5\!+ \hdots +\!q) \:=\: (1\!+\!3\!+\!5\!+ \hdots +\!33)$

Fact: .The sum of the first $n$ odd numbers is $n^2.$

We have:

. . $\underbrace{(1\!+\!3\!+\!5\!+ \hdots +\!p)}_{\frac{p+1}{2}\text{ terms}} + \underbrace{(1\!+\!3\!+\!5\!+ \hdots +\!q)}_{\frac{q+1}{2}\text{ terms}} \:=\: \underbrace{(1\!+\!3\!+\!5\!+ \hdots +\!33)}_{\text{17 terms}}$

Hence: . $\left(\dfrac{p+1}{2}\right)^2 + \left(\dfrac{q+1}{2}\right)^2 \;=\;17^2$

We have a Pythagorean Triple: . $a^2 + b^2 \:=\:17^2$

. . and the only solution is: . $\{a,b\} \:=\:\{8,15\}$

So we have: . $\begin{Bmatrix}\dfrac{p+1}{2} &=& 8 && \Rightarrow && p &=& 15 \\ \\[-3mm] \dfrac{q+1}{2} &=& 15 && \Rightarrow && q &=&29 \end{Bmatrix}$

Therefore: . $p + q \:=\:15 + 29 \:=\:44$