4it is a polynomial of degree (11) (!2)/2 = 66
So to get coeff of x59 we get - (sum of all the roots taken 7 at a time)
= - (1 + 2√2 +3 3√ 3 + 41/4 ) + ..........
Hope u got it
Find the co-efficient of x59 in the expansion :
(x-1)(x2-2)(x3-3)(x4-4).........(x11-11)
EXPERT'S HELP PLEASE !!!
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5 Answers
4
341Every term in the expansion is obtained by choosing terms from every bracket, either the constant term or the power of x and multiplying them out.
If from every bracket we choose the power of x, we get x66.
So to obtain x59 we must ignore, from some brackets, distinct powers of x adding up to 7 and choose the constant term appearing in those brackets. The sum of the contributions from the constant terms will be the coefficient we are looking for
Hence the different ways to get x59 are:
1) by ignoring x7. The contribution is -7.
2) by ignoring x1 and x6. This gives (-1) X (-6) = 6.
3) x2 and x5 which gives 10
4) x3 and x4 which gives 12
5) x,x2 and x4 which gives -8
So the coefficient = -7+6+10+12-8 = 13
4@prophet Sir I understood ure solution
But y am i wrong
I considered that as polynomial of degree 66 which has roots 1 , √2 ,3√3,41/4......
Where's the mistake ??????????
341Its not that you are wrong. But aveek wants a definite number, which is difficult to arrive at by this process. So, you will have to use a different method, such as the one above, to evaluate the coefficient.
Also by taking roots seven at a time, you have choose seven of them, multiply them out and add them. Thats quite long drawn out, you agree?
