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If m,n,r are positive integers such that...

If m,n,r are positive integers such that r `lt ` m,n, then
`""^(m)C_(r)+""^(m)C_(r-1)""^(n)C_(1)+""^(m)C_(r-2)""^(n)C_(2)+...+ ""^(m)C_(1)""^(n)C_(r-1)+""^(n)C_(r)` equals

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Here , sum of lower suffixes of binomial cofficients in each term is r.
i.e. ` r = r- 1 + 1 = r - 2 + 2 = …= r = r`
Since ,
`(1 + x)^(m)= ""^(m)C_(0) + ""^(m)C_(1)x + ...+ ""^(m)C_(r-2)x^(r-2) + ""^(m)C_(r-1) x^(r-1) + ""^(m)C_(r) x^(r) + ...+ ""^(m)C_(m) x^(m) `...(i)
and ` (1 + x)^(n) = ""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2)x^(2) + ...+ ""^(n)C_(r)x^(r) + ...+ ""^(n)C_(n) x^(n) ` ...(ii)
On multiplying Eqs .(i) and (ii) , we get
`((1 + x)^(m+n)= ""^(m)C_(0) + ""^(m)C_(1)x + ...+ ""^(m)C_(r-2)x^(r-2) + ""^(m)C_(r-1) x^(r-1)`
` + ""^(m)C_(r) x^(r) + ...+ ""^(m)C_(m) x^(m))xx(""^(n)C_(0)+ ""^(n)C_(r) x+ ""^(n)C_(2)x^(2)`
` + ...+ ""^(n)C_(r) x^(r) + ...+ ""^(n)C_(n) x^(n))...`(iii)
Now , coefficient of ` x^(r)` in RHS
`= ""^(m)C_(r) * ""^(n)C_(0) + ""^(m)C_(r-1) ""^(n)C_(1) + ""^(m)C_(r-1) *""^(n)C_(2) + ...+ ""^(m)C_(0) * ""^(n)C_(r) `
` ""^(m)C_(r) + ""^(m)C_(r-1) ""^(n)C_(1) * ""^(m)C_(r-2) * ""^(n)C_(2) + ...+ ""^(n)C_(r)`
Coefficient of ` x^(r) ` in LHS ` = "" ^(m +n)C_(r)`
But Eq.(iii) is an identity , therefore coefficient of ` x^(r) ` in LHS
=coefficient of ` x^(r)` in RHS
` rArr ""^(m+n)C_(r) = ""^(m)C_(r) + ""^(m)C_(r-1) * ""^(n)C_(1) + ""^(m)C_(r-2) * ""^(n)C_(2) + ...+ ""^(n)C_(r)` .
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ARIHANT MATHS-BIONOMIAL THEOREM-Exercise (Questions Asked In Previous 13 Years Exam)
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