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The coefficient of x^7 in expansion of (1-x-x^2+x^3)^6 is
Dear Niharika (1 - x - x2 + x3)6= (1 - x)6(1 - x2)6= (6C0 - 6C1 x + ... + 6C6 x6)(6C0 - 6C1 x2 + ... + 6C6 x12)Now, coefficient of x7 is = 6C1 6C3 - 6C3 6C2+ 6C5 6C1 = – 144RegardsArun (askIITians forum expert)
(1 - x - x2 + x3)6
= (1 - x)6(1 - x2)6
= (6C0 - 6C1 x + ... + 6C6 x6)(6C0 - 6C1 x2 + ... + 6C6 x12)
Now, coefficient of x7 is = 6C1 6C3 - 6C3 6C2+ 6C5 6C1
= – 144
Regards
Arun (askIITians forum expert)
The answer given above may be wrong, the correct method is u can write the cubic into factors and they are as follows (1-x)^7(1+x)^6 now if u write the general term of both and they are as follows [7Cm.(-1)^m.(x) ^m]. [6Cn.(x)^n] now u can compare as we want that m+n=7 so total pairs possible are (m, n) = (7,0), (6,1), (5,2), (4,3), (3,4), (2,5), (1,6) now if u put value in the formula one by one then u will be getting this result as follows [-1+42-315+700-525+126-7]= 20
Dear student,Please find the attached solution to your problem. (1 - x - x2 + x3)6 = (1 - x)6(1 - x2)6= (6C0 - 6C1 x + ... + 6C6 x6)(6C0 - 6C1 x2 + ... + 6C6 x12) Now, coefficient of x7 is = 6C1 6C3 - 6C3 6C2+ 6C5 6C1= – 144 Hope this helps.Thanks and regards,Kushagra
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