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1.

According to Brooks, if n is number of programmers in a project team the number of communication paths is:1. n (n - 1)/22. n log n3. n4. n (n + 1)/2

Answer» Correct Answer - Option 1 : n (n - 1)/2

Brooks introduced a formula to measure the volume of communication within the team. It assumes that every team member needs to communicate with every other team member, in order to perform his or her work.  

Group intercommunication formula measures the number of channels of communication within the project team and can be expressed as n*(n-1)/2, where n is the number of people working on a project
Discussion
2.

Consider the following two sequences,X = {-1, 1, -1, 1, ...} andY = {1, \(\frac{1}{2}\), 3, \(\frac{1}{4}\),...}.Then:1. both X and Y are divergent.2. both A and Y are convergent.3. X is convergent but Y is divergent.4. X is divergent but Y is convergent.

Answer» Correct Answer - Option 1 : both X and Y are divergent.

Concept:

Let, an is the nth term of a sequence then if,

\(\mathop {\lim }\limits_{n \to ∞ } {a_n} = 0\)

then the sequence converges else the sequence diverges

Analysis:

<!--[if gte vml 1]><![endif]--><!--[if !vml]-->

\(X = \left\{ { - 1,\;1,\; - 1,\;1,\; \ldots } \right\}\)

an = (-1)n

<!--[if gte msEquation 12]>limn→∞an=-1∞=1≠0<![endif]--><!--[if !msEquation]--><!--[if gte vml 1]><![endif]--><!--[if !vml]--><!--[endif]--><!--[endif]-->\(\mathop {\lim }\limits_{n \to ∞ } {a_n} = {\left( { - 1} \right)^∞ } \ne 0\) 

∴ X is divergent

<!--[if gte vml 1]><![endif]--><!--[if !vml]-->\(Y = \left\{ {1,\frac{1}{2},\;3,\frac{1}{4}, \ldots } \right\}\)<!--[endif]--><!--[endif]-->

Now,

Yn = n; if n = odd

= 1/n; if n = even

For even terms the sequence is convergent (Since 1/n tends to 0 as n tends to ∞ )

For odd terms the sequence is divergent. 

∴ We cannot say that the sequence is convergent. 
Discussion
3.

The p-series \(\mathop \sum \nolimits_{n = 1}^\infty \frac{1}{{{n^p}}}\) diverges for:1. 0 14. p ϵ [2,4,6,.......]

Answer» Correct Answer - Option 1 : 0 <  p ≤ 1

Concept:

A p-series is a specific type of infinite series. It's a series of the form as shown below,

\(\mathop \sum \nolimits_{n = 1}^\infty \frac{1}{{{n^p}}} = \frac{1}{{{1^p}}}+\frac{1}{{{2^p}}}+\frac{1}{{{3^p}}}+... \)

where p can be any real number greater than zero.

Notice that in this definition n will always take on positive integer values, and the series is an infinite series because it's a sum containing infinite terms.

There are infinitely many p-series because you have infinite choices for p. Each time you choose a different value for p you create another p-series.

With p-series,

If p > 1, the series will converge, or in other words, the series will add up to a specific numerical value.

If 0 < p ≤ 1, the series will diverge, which means that the series won't add up to a specific numerical value.
Discussion
4.

The sum of the series 1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ........... will be:1. \(\frac 1 {a^4}\)2. 13. \(-\frac 1 {a^2}\)4. -1

Answer» Correct Answer - Option 1 : \(\frac 1 {a^4}\)

Concept:

a + ar + ar2 + ar3 +….. 

Sum of the above infinite geometric series:

\(=\frac{a}{1-r}\)

Analysis:

Given:

1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ......

let x = (a2 + 1)

The series now becomes

S = 1 + 2x + 3x2 + 4x3 + ......  ----(1)

By multiplying x on both sides we get

xS = x + 2x2 + 3x3 + 4x4 + ...... ----(2)

Subtracting (1) and (2), we get

S(1 - x) = 1 + x + x2 + x3 + ..... ---(3)

The right hand side of (3) forms infinite geometric series with a = 1, r = x

∴ S(1 - x) = \(\frac{1}{1-x}\)

\(\Rightarrow S = \frac{1}{(1-x)^2}\)

putting the value of x, we get

\(\Rightarrow S = \frac{1}{(1- a^2 - 1)^2}\)

\(\Rightarrow S = \frac{1}{a^4}\)
Discussion