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Problem Of The Week #426 July 20th, 202

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anemone

MHB POTW Director
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Feb 14, 2012
3,632
Here is this week's POTW:

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Assume that $x_1,\,x_2,\,\cdots,\,x_7$ are real numbers such that

$x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1\\4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12\\9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123$

Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,632
Congratulations to kaliprasad for his correct solution, which you can find below:

Solution from kaliprasad :
We recognize that the 3 given equations are

$\sum_{n=1}^7n^2x_n= 1\cdots(1)$

$\sum_{n=1}^7(n+1)^2x_n= 12\cdots(2)$

$\sum_{n=1}^7(n+2)^2x_n= 123\cdots(3)$

And we need to evaluate $\sum_{n=1}^7(n+3)^2x_n$

Now let us formulate a result that we shall use

We have $(y+1)^2 - y^2 = 2y+1$

Using this for y = m and y = m+1 we get 2 relations

$(m+1)^2 - m^2 = 2m + 1\dots(4)$

$(m+2)^2 - (m+1)^2 = 2m + 3\cdots(5)$

Subtract (4) from (5) to get $(m+2)^2 + m^2 - 2(m+1)^2 = 2\cdots(6)$

Putting m+1 in place of m we get (as above is true for any m )

$(m+3)^2 + (m+1)^2 - 2(m+2)^2 = 2\cdots(7)$

From (6) and (7) we get

$(m+3)^2 + (m+1)^2 - 2(m+2)^2 = (m+2)^2 + m^2 - 2(m+1)^2 $

Or
$(m+3)^2 = 3(m+2)^2 - 3 (m+1)^2 + m^2\cdots(8)$

Now
$\sum_{n=1}^7(n+3)^2x_n$

= $\sum_{n=1}^7(3(n+2)^2 - 3 (n+1)^2 + n^2)x_n$ (using (8))

$= 3\sum_{n=1}^7(n+2)^2x_n - 3 \sum_{n=1}^7(n+1)^2x_n + \sum_{n=1}^7n^2x_n$

$= 3 * 123 - 3 * 12 + 1 = 334$ (using given conditions(1), (2), (3) )
 
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