यदि `(If)""^(k+5)P_(k+1) = (11(k-1))/(2) । ""^(k+3)P_(k)` ज्ञात कीजिए ।

If \ ^(k+5)P_(k+1)=(11(k-1))/2dot\ ^(k+3)P_(k\ )P_k then the values of k are 7\ a n d\ 11 b. 6\ a n d\ 7 c. 2\ a n d\ 11 d. 2\ a n d\ 6

If ^(k+5)P_(k+1)=(11(k-1))/(2)*k+3P_(k)P_(k) then the values of k are 7 and 11 b.6 and 7c.2 and 11 d.2 and 6

60.A consecutive reaction occurs with two equilibria which co-exist together P(k_(1))/(k_(2) Q (k_(3))/(k_(4) R Where k_(1) , k_(2) , k_(3) and k_(4) are rate constants.Then the equilibrium constant for the reaction P harr R is (1) (k_(1)*k_(2)) / (k_(3)*k_(4) ) (2) (k_(1)*k_(4)) / (k_(2)k_(3) ) (3) k_(1) * k_(2) * k_(3)*k_(4) (4) (k_(1)*k_(3)) / (k_(2)k_(4)) ]]

If P(10, k+1): P(11, k) = 30 : 11, then

If P (n) = 2 + 4 + 6+…+ 2n, n in N then P(k) = k (k +1) + 2 implies P (k +1 ) = (k +1 ) (k + 2)+ 2 AA k in N. So, we can conciude that P (n) = n (n +1) + 2 for

The p.m.f. of a.r.v. X is as follows: P( X=0) = 3k^ (3), P(X=1) = 4k-10k^ (2) , P(X=2)=5k-1, P(X=x)=0 for any other value of x, then value of k is

The p.m.f. of a.r.v. X is as follows: P( X=0) = 3k^ (3), P(X=1) = 4k-10k^ (2) , P(X=2)=5k-1, P(X=x)=0 for any other value of x, then value of k is

If the value of the sum n^(2) + n - sum_(k = 1)^(n) (2k^(3)+ 8k^(2) + 6k - 1)/(k^(2) + 4k + 3) as n tends to infinity can be expressed in the form (p)/(q) find the least value of (p + q) where p, q in N