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Valentas

08-08-2011, 05:02 AM

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?

galactus

08-08-2011, 08:54 AM

Try setting up the general sum for the positive terms, then use the formula for summing of integers, squares, etc.

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

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

Generally, we have \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.

Denis

08-08-2011, 10:57 AM

HINT: positive terms:

8 48 120 224 360 .....

40 72 104 136 .....

32 32 32 .....

Valentas

08-08-2011, 12:02 PM

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 :)

galactus

08-08-2011, 12:21 PM

Then let's do this one.

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

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

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

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

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

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

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

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

So, we get:

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

This is subtracted from the other sum:

\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:

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.

Subhotosh Khan

08-08-2011, 12:26 PM

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.......

Subhotosh Khan

08-08-2011, 09:14 PM

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

Valentas

08-16-2011, 06:30 AM

= (-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) .

Subhotosh Khan

08-16-2011, 08:32 AM

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 ?

Valentas

08-16-2011, 08:40 AM

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. :(

Subhotosh Khan

08-16-2011, 08:45 AM

Elaborate please. I can't see anything special in the line above. :(

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

Valentas

08-16-2011, 09:11 AM

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?

Subhotosh Khan

08-16-2011, 10:36 AM

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....

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