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I am stuck in this maths problem. Please somebody help:
Exercise 9d Q2(d) of this book:
http://books.google.com.pk/books?id=gB5 ... &q&f=false
I found the Σ(a^2) and Σ(1/a) successfully correct bt culdn't find Σ(a^2)(B^2) and (Σa)(ΣaB):
'ΣaB(a + B) = 6' is given as the answer but my own answer comes 4. Please somebody post the complete solution for ΣaB(a + B) explaining each step, given that i found ΣaB(a + B) = (Σa)(ΣaB) - 2(ΣaBc).
It seems i should have got ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc) in order to get the correct answer '6' instead of the erroneous '4'.
If perhaps this is true, then prove that ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc)
I found Σ(a^2)(B^2) = (ΣaB)^2 - 2Σ(a.B^2.C) - 4aBCD. Now 4m here i can't proceed 4ward 2 find Σ(a.B^2.C).
NB: a,B,C,D represent alpha, beta, gamma and the 4th root (partial derivative sign) respectively.
Exercise 9d Q2(d) of this book:
http://books.google.com.pk/books?id=gB5 ... &q&f=false
I found the Σ(a^2) and Σ(1/a) successfully correct bt culdn't find Σ(a^2)(B^2) and (Σa)(ΣaB):
'ΣaB(a + B) = 6' is given as the answer but my own answer comes 4. Please somebody post the complete solution for ΣaB(a + B) explaining each step, given that i found ΣaB(a + B) = (Σa)(ΣaB) - 2(ΣaBc).
It seems i should have got ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc) in order to get the correct answer '6' instead of the erroneous '4'.
If perhaps this is true, then prove that ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc)
I found Σ(a^2)(B^2) = (ΣaB)^2 - 2Σ(a.B^2.C) - 4aBCD. Now 4m here i can't proceed 4ward 2 find Σ(a.B^2.C).
NB: a,B,C,D represent alpha, beta, gamma and the 4th root (partial derivative sign) respectively.