progression

Valentas

New member
Joined
Dec 19, 2010
Messages
34
I just started to do some maths because it is going to be my last year and I met this problem:

2 · 4 - 4 · 6 + 6 · 8 - 8· 10 + ... + 98 · 100 - 100 · 102 =?

I completely forgot how to solve these. Could anyone remind me?
 
Try setting up the general sum for the positive terms, then use the formula for summing of integers, squares, etc.

i.e \(\displaystyle \displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\)

The positive terms: \(\displaystyle 2\cdot4 +6\cdot 8+10\cdot 12+\cdot\cdot\cdot +98\cdot 100\)

Generally, we have \(\displaystyle \displaystyle\sum_{k=1}^{25}(4k-2)4k=8\displaystyle\sum_{k=0}^{25}(2k^{2}-k)\)

Now, do the same for the negative terms. Then, subtract this total from the positive terms.
 
Is this the only way for this to solve? Because I finished 11 grades only and we didn't learn about integers. Any progressions pattern? Geometric,arithmetic? I can't see the crux on this problem because the summer has washed my knowledge :)
 
Then let's do this one.

The negative terms are \(\displaystyle 4k(4k+2)\)

See it?. Put in k=1 and we get \(\displaystyle 4\cdot 6\)

Put in a k=2 and we get \(\displaystyle 8\cdot 10\) and so on.

Sum these all up. Noting that the last term corresponds to k=25

\(\displaystyle \sum_{k=1}^{25}4k(4k+2)\)

\(\displaystyle \sum_{k=1}^{25}(16k^{2}+8k)\)

Note that \(\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}\) and

\(\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}\)

So, we get:

\(\displaystyle 16\cdot\frac{n(n+1)(2n+1)}{6}+8\cdot\frac{n(n+1)}{2}\)

This is subtracted from the other sum:

\(\displaystyle \sum_{k=1}^{n}(16k^{2}-8k)=16\cdot\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{n(n+1)}{2}\)

The only difference between the two is the sign in the middle.

So, we get:

\(\displaystyle 16\cdot\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{n(n+1)}{2}-\left(16\cdot\frac{n(n+1)(2n+1)}{6}+8\cdot\frac{n(n+1)}{2}\right)=-8n(n+1)\)

Plug in n=25 to get the total.
 
I just started to do some maths because it is going to be my last year and I met this problem:

2 · 4 - 4 · 6 + 6 · 8 - 8· 10 + ... + 98 · 100 - 100 · 102 =?

I completely forgot how to solve these. Could anyone remind me?

2 * 4 - 4 * 6 + 6 * 8 - 8* 10 + 10*12 - 12*14 + 14*16 - 16*18 ... + 98 · 100 - 100 · 102

= [ 2 * 4 - 4 * 6 ]+ [6 * 8 - 8* 10] + [10*12 - 12*14] + [14*16 - 16*18] ... + [98 · 100 - 100 · 102]

= 4*(-4) + 8*(-4) + 12*(-4) .... 100*(-4)


and continue.......
 
2 * 4 - 4 * 6 + 6 * 8 - 8* 10 + 10*12 - 12*14 + 14*16 - 16*18 ... + 98 · 100 - 100 · 102

= [ 2 * 4 - 4 * 6 ]+ [6 * 8 - 8* 10] + [10*12 - 12*14] + [14*16 - 16*18] ... + [98 · 100 - 100 · 102]

= 4*(-4) + 8*(-4) + 12*(-4) .... 100*(-4)

= (-16)*[1 + 2 + 3 + 4 + ....+23 + 24 + 25]

= (-16) * 25 * 26/2

= - 5200
 
Hi I just came back from holiday. I understand everything except where you get 25 *26/2 after (-16) .

Compare with the statement made in the line above.

What does it equate to?

Do you know the expression for the series = 1 + 2 + 3 + 4 ..... +(n-2) + (n-1) + n ?
 
Compare with the statement made in the line above.

What does it equate to?

Do you know the expression for the series = 1 + 2 + 3 + 4 ..... +(n-2) + (n-1) + n ?

Elaborate please. I can't see anything special in the line above. :(
 
It seems like no. Could you remind me and I will finish.

Maybe I understood.

{1 + 2 + 3 + ,,, + 25}

Brackets are arithmetic progression I think. Because d = 1 .

I can write it in S(25)=a1+a25/2 *25 ? Then I get 1 + 25/2 * 25 = 13 * 25 = 325

325 * (-16) = -5200 ! OK?
 
Last edited:
It seems like no. Could you remind me and I will finish.

Maybe I understood.

{1 + 2 + 3 + ,,, + 25}

Brackets are arithmetic progression I think. Because d = 1 .

I can write it in S(25)=a1+a25/2 *25 ? Then I get 1 + 25/2 * 25 = 13 * 25 = 325

325 * (-16) = -5200 ! OK?

You got it....
 
Top