ISI 2010

[rongmon]

19.
Consider any finite integer r ≥ 2. Then lim(x→0) f(r, x)/a(x) equals,
(where f(r, x) = log(e)(Σ(0, r) (x^k)); and a(x) = Σ(1, ∞) ((x^k)/k!))
(a) 0,
(b) 1,
(c) e,
(d) log(e)2.

Note: Σ(a, b) g(k) is summation of g(k) over values of k from a to b
x^k is x to the power k

[rongmon]

17.

Let Xi, i = 1, 2, . . . , n be identically distributed with variance S^2. Let
cov(Xi,Xj) =  P for all i not equal to j.

Define Xn = (1/n) * Summation Xi and let An = Var(Xn).

Then lim of An as n tends to infinity equals
(a) 0,
(b) P,
(c) S^2 + P,
(d) S^2 + P^2.

[Noel]

B for 19
the numerator would be
log(1-x^(r+1))-log(1-x)
and the denominator would be
e^x-1
use l'hopital's rule

[rongmon]

Question 21.

Thank you.

[rongmon]

Thanks Noel.

[rongmon]

Did you simplify denominator using sequence summation?

[Noel]

I didn't get you rongmon..well summation of  x^k/k! where k varies from 0 to infinity is e^x hence if k varies from 1 to infinity then the value becomes e^x-1

[rongmon]

Oh okay, thanks.

Is there a formula for such sequences?

[Tania123]

Can someone explain Q21 please?

[samuraijack]

@Tania

I guess answer should be D
use nCr + nCr-1 = n+1Cr

(nC0 + n+1C1 ) + n+2C2......n+mCm

(n+1C0 + n+1C1) + n+2C2......+ n+mCm

(n+2C1) + n+2C2......+ n+mCm

(n+2C1 + n+2C2)+........n+mCm
(n+3C2) + n+3C3+.....n+mCm
Finally
n+mCm-1 + n+mCm = n+m+1Cm= n+m+1Cn+1 Option D