33Ek tarika to hai ki... if n=1 or n=2 to ye zero hoga... aur n=3 par 1
tahts why....
:D :D
serious wala expalanation de rahe hain... :P
\sum_{1\ll i }^{}{}\sum_{<j<k}^{}{}\sum_{\ll n}^{}{(x_{i}+x_{j}+x_{k}})=\lambda (\sum_{m=1}^{n}{x_{m}})
Determine λ
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17 Answers
33
24ya good work priyam.............maine no. of terms galat gin liye.......[2][2]..isiliye ans nahin aaya.............thanx.......[1]
33and λ=no of x1 = no x2 = no of x3 = no of x4.....
as RHS =λ(x1+x2+x3+...+xn)
33Expanding the summation...
we get..
(x1+x2+x3)+(x1+x2+x4)+(x1+x2+x5).....+(x1+x2+xn)+... (n-2 terms)
+(x1+x3+x4)+(x1+x3+x5)+......+(x1+x3+xn)+...(n-3 terms)
....
...
+(x1+xn-1+xn).(n-(n-1) terms)+.....+.....(Rest terms don't have x1)
total no of x1=(n-2)+(n-3)+...+(n-(n-1))
=(n-1)(n-2)/2
