So many summations...........

\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

17 Answers

33
Abhishek Priyam ·

Ek 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

24
eureka123 ·

ya good work priyam.............maine no. of terms galat gin liye.......[2][2]..isiliye ans nahin aaya.............thanx.......[1]

1
skygirl ·

nice priyam :)

21
tapanmast Vora ·

Yeah! Questn aur Ans dono!!!

thnx [1]

33
Abhishek Priyam ·

Ye kya hai... :D

same copy kar diya.. :P

33
Abhishek Priyam ·

@tapan tumko q samajh me aaya ab... ??

33
Abhishek Priyam ·

and λ=no of x1 = no x2 = no of x3 = no of x4.....

as RHS =λ(x1+x2+x3+...+xn)

33
Abhishek Priyam ·

Expanding 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

33
Abhishek Priyam ·

Q me 1≤i<j<k≤n

hoga na ki..

1<<i<j<k<<n

1
Akand ·

is there no relation between i j k n m????????? man its difficult

21
tapanmast Vora ·

DUDE can u first xplain the questn [2]

24
eureka123 ·

yup....................explain.............[1][1]

33
Abhishek Priyam ·

waise is the answer

\frac{(n-1)(n-2)}{2}

33
Abhishek Priyam ·

:D

24
eureka123 ·

big brother of double summation[9]

24
eureka123 ·

triple summation.....[1]

21
tapanmast Vora ·

yeh kya bhai [7]

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