A book of mine has the following questions with no answers.So I am not sure if I am right or if the proceure is correct and the best.So,plz help-(adsbygoogle = window.adsbygoogle || []).push({});

1.For a given real number a>0,define[tex] A_n=(1^a + 2^a + 3^a +....+n^a)^n[/tex]

and[tex] B_n=n^n(n!)^a[/tex] for all n=1,2,.... Then

a. [tex]A_n<B_n[/tex]for all n>1

b.there exists an integer n>1 such that [tex]A_n<B_n[/tex]

c.[tex]A_n>B_n[/tex] for all n>1

d.there exists integers n and m both larger than one such that [tex]A_n>B_n[/tex] and [tex]A_m<B_m[/tex]

As there's no specification about the value of a in the options given,I considered a special case taking a=1 and then put n=2 and n=3.I found [tex]A_n>B_n[/tex] and [tex]A_n<B_n[/tex] respectively.So I think the answer is d.

2.Is 1 a prime number?

3.Let C denote the set of all complex numbers.Define A and B by

A={(z,w):z,w [tex]\in [/tex]C and mod z=mod w}

B={(z,w):z,w [tex]\in [/tex]C and [tex]z^2=w^2[/tex]}

Then

a.A=B b.A[tex]\sqsubseteq[/tex]B and A not equal to B

c.B[tex]\sqsubseteq[/tex]A and B not equal to A

d.none of the above.

I think z^2=w^2 means mod z=mod w but the reverse is not true,so is the answer b?

4.If positive numbers a,b,c,d are such that 1/a,1/b,1/c,1/d are in A.P then we always have

a.a+d[tex]\geq[/tex]b+c b.a+b[tex]\geq[/tex]c+d

c.a+c[tex]\geq[/tex]b+d d.none of the above

1/a +1/d =1/b +1/c or,(a+d)/ad =(b+c)/bc. Now 1/ad<1/bc or,ad>bc.so,a+d[tex]\geq[/tex]b+c i.e.,(a)?

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thanking you in advance.And how do you put "not equal to" and "modulus" in latex?

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# Homework Help: Plzz check the answers to these difficult questions?

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