Either 1000 or a constant of variation
11 + k
That depends on the value assigned to the variable "k". If you don't know the value of "k", you express the sum as 11 + k; in this case, you can't simplify it any more.
k12
(9+k)-17=x
100/(k+m)Its actual value depends on the values of 'k' and 'm' .
17
The sum of the even numbers up to 2k, where k is an integer, is k(k + 1) = k2 + k
The flowchart to read 10 positive integers K>10 Start A N K=1 Sum = 0 Sum = Sum + K2 B Is Y Print K > 100? sum K=k+1 End B A
11 + k
k + 7
Print "Type the upper limit (n) ?" Input n K = -1 WHILE K < = n K = K + 2 Sum = Sum + K WEND Print "The sum of all odd numbers up to "; n; "is "; Sum
enter the number whose digits are to be added num is the given value num=0! k=num%10 sum=sum=k k=num/10 num=k print the sum of the digits
it would have a part in it like this: for (i=0; i<n; ++i) { . for (j=0; j<l; ++j) { . . sum= 0; . . for (k=0; k<m; ++k) { . . . sum += a[i][k] * b[k][j]; . . } . . c[i][j] = sum; . } }
2(k+4)
k^2 + k = k^2 + k k^2 x k = k^3
If X has the Poisson distribution with mean l then Pr(X = k) = e-llk/k! Mean of Poisson = Sum over all k of [k*P(X = k)] which happens to be l. = Sum over all k of [k*e-llk/k!] = Sum over all k of [e-llk/(k-1)!] = Sum over all j of [le-llj/j!] where j has been substituted for k-1 = l*Sum over all j of [e-llj/j!] But the quantity being summed is simply the pdf of the Poisson distribution and so its sum over all possible values is 1 So Mean = l And then Variance of Poisson = Sum over all k of [k2*P(X = k)] - l2. = Sum over all k of [k2*e-llk/k!] - l2 Then, since k2 = k*(k-1) + k Variance = Sum over all k of [k*(k-1)e-llk/k!] +Sum over all k of [k*e-llk/k!] - l2 = Sum over all j of [l2e-llj/j!] where j has been substituted for k-2 + Sum over all i of [le-lli/i!] where i has been substituted for k-1 - l2 = l2*Sum over all j of [e-llj/j!] + l*Sum over all i of [e-lli/i!] - l2 And since the sums are equal to 1, Variance = l2 + l - l2 = l Apologies: I did the answer using the symbol for lambda but this browser changed them all back to l. I cannot change them all t o something else but I hope it is clear. At least they are distinguishable from 1!
That depends on the value assigned to the variable "k". If you don't know the value of "k", you express the sum as 11 + k; in this case, you can't simplify it any more.