JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

Let

n( > 1)

be a positive integer then the largest integer m such that

\left( {{n^m} + 1} \right)

divides

\left( {1 + n + {n^2} + ... + {n^{127}}} \right)

, is

View answer and explanation

Correct option: C

Option C (64) is correct. The expression

\frac{{1 + n + {n^2} + ... + {n^{127}}}}{{{n^m} + 1}}

is an integer when

m = 64

because it allows for factorization using the difference of squares.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}