The sum of the series $1/(1 + {1^2} + {1^4}) + 1/(1 + {2^2} + {2^4}) + 1/(1 + {3^2} + {3^4}) + ...$ to $n$ terms is

$\frac{{n\left( {{n^2} + 1} \right)}}{{{n^2} + n + 1}}$

$\frac{{n\left( {{n^2} - 1} \right)}}{{2({n^2} + n + 1)}}$

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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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841 Easy

841 of 998

Analyzing +4 | -1

Number of real roots of equation

3^{\log _3\left(x^2-4 x+3\right)}=(x-3)

View answer and explanation

Correct option: A

Option A (0) is correct because after solving the equation, the roots obtained do not satisfy the domain of the logarithmic function, resulting in no real roots.

842 Easy

842 of 998

Analyzing +4 | -1

If both the regression coefficients

b_{y x}

and

b_{x y}

are positive, then

View answer and explanation

Correct option: A

\frac{1}{b_{y x}}+\frac{1}{b_{x y}} > \frac{2}{r}

Option A (

\frac{1}{b_{y x}}+\frac{1}{b_{x y}} > \frac{2}{r}

) is correct because the geometric mean of regression coefficients is greater than or equal to the harmonic mean.

843 Easy

843 of 998

Analyzing +4 | -1

x = \sum \limits_{n = 0}^\infty {a^n},y = \sum \limits_{n = 0}^\infty {b^n},z = \sum \limits_{n = 0}^\infty {c^n},

where

a,b,c

are in A.P. such that

\left| a \right| < 1,\left| b \right| < 1,

and

\left| c \right| < 1,

then

x,y,z

are in

View answer and explanation

Correct option: C

H.P.

Option C (H.P.) is correct because the reciprocals of x, y, and z are in A.P., implying that x, y, and z are in H.P.

844 Easy

844 of 998

Application +4 | -1

p

th term of an arithmetic progression is

q

and the

q

th term is

p

, then 10th term is

View answer and explanation

Correct option: D

p + q - 10

Option D is correct because by solving the system of equations formed by the given conditions, we can find the first term and common difference, and then determine the 10th term.

845 Easy

845 of 998

Application +4 | -1

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is

View answer and explanation

Correct option: B

2+\sqrt{3}

Option D (

\frac{{1}}{{26}}(3^{50}-1)

) is correct because it is the sum of the first 50 terms of the G.P. calculated using the given conditions.

846 Easy

846 of 998

Application +4 | -1

\tan n\mathrm{\theta } = \tan m\mathrm{\theta },

then the different values of

\mathrm{\theta }

will be

View answer and explanation

Correct option: A

Option A (AP) is correct because the different values of

\theta

form an arithmetic progression.

847 Easy

847 of 998

Application +4 | -1

The value of

\frac{{\log 49\sqrt 7 + \mathrm{log}25\sqrt 5 - \log 4\sqrt 2 }}{{\log 17.5}}

View answer and explanation

Correct option: C

\frac{5}{2}

Option C (

\frac{5}{2}

) is correct because simplifying the logarithmic expression results in

\frac{5}{2}

848 Easy

848 of 998

Analyzing +4 | -1

\log _5 120+(x-3)-2 \log _5\left(1-5^{x-3}\right)

=-\log _5\left(0.2-5^{x-4}\right),

then

x

View answer and explanation

Correct option: A

Option A (1) is correct because solving the logarithmic equation leads to x = 1.

849 Easy

849 of 998

Analyzing +4 | -1

S = \sum \limits_{n = 1}^\infty \left( {\frac{{{\;^n}{C_0} + {\;^n}{C_1} + {\;^n}{C_2} + \ldots + {\;^n}{C_n}}}{{{\;^n}{P_n}}}} \right),

then

S

equals

View answer and explanation

Correct option: D

None of these

Option C (Both (a) and (b)) is correct because, given that a, b, c, d are in HP, both the relations

a + d > b + c

and

ad > bc

can be derived.

850 Easy

850 of 998

Application +4 | -1

The sum of the series

\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\cdots

to 9 terms is

View answer and explanation

Correct option: D

-\frac{3}{2}

Option D (

-\frac{3}{2}

) is correct because the sum of the series to 9 terms is calculated using the AP sum formula, resulting in

-\frac{3}{2}

851 Easy

851 of 998

Application +4 | -1

{\log _6}\left( {x + 3} \right) - {\log _6}x = 2,

then

x =

View answer and explanation

Correct option: B

\frac{3}{{35}}

Option B (

\frac{3}{{35}}

) is correct because it is found by solving the logarithmic equation using properties of logarithms.

852 Easy

852 of 998

Analyzing +4 | -1

Let

{S_n} = \frac{1}{{{1^3}}} + \frac{{1 + 2}}{{{1^3} + {2^3}}} + ... + \frac{{1 + 2 + ... + n}}{{{1^3} + {2^3} + ... + {n^3}}},\;n = 1,2,3, \ldots

Then

{S_n}

is not greater than

View answer and explanation

Correct option: C

Option C (2) is correct because the sum of the series

{S_n}

is always less than or equal to 2, as shown by simplifying the general term and summing the series.

853 Easy

853 of 998

Analyzing +4 | -1

The sum of the series

\frac{1^2}{1 \cdot 2 !}+\frac{1^2+2^2}{2 \cdot 3 !}+\frac{1^2+2^2+3^2}{3 \cdot 4 !}+\ldots+\frac{1^2+2^2+\ldots+n^2}{n \cdot(n+1) !}+\ldots \infty

: Equals

View answer and explanation

Correct option: C

\frac{3 e-1}{6}

Option C ((3e-1)/6) is correct because the sum of the given infinite series can be simplified and expressed in terms of the exponential constant 'e'.

854 Easy

854 of 998

Analyzing +4 | -1

The product

2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \ldots

\infty

is equal to:

View answer and explanation

Correct option: A

2^{\frac{1}{2}}

Option A (

2^{\frac{1}{2}}

) is correct because by expressing each term as a power of 2 and summing the exponents, we obtain a G.P. with a sum of 1/2, resulting in

2^{\frac{1}{2}}

855 Easy

855 of 998

Analyzing +4 | -1

The limiting value of the sum to

n

terms of the series

\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots \ldots \ldots

n \to \infty ,\,

View answer and explanation

Correct option: D

Option D (1) is correct because the limiting value of the sum is found by expressing the general term as a difference and then evaluating the limit of the telescoping series.

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856 Easy

856 of 998

Analyzing +4 | -1

{S_1},{S_2}

and

{S_3}

denote the sum of first

{n_1},{n_2}

and

{n_3}

terms respectively of an A.P., then

\frac{{{S_1}}}{{{n_1}}}\left( {{n_2} - {n_3}} \right) + \frac{{{S_2}}}{{{n_2}}}\left( {{n_3} - {n_1}} \right) + \frac{{{S_3}}}{{{n_3}}}\left( {{n_1} - {n_2}} \right) =

View answer and explanation

Correct option: A

Option A (0) is correct because the given expression simplifies to zero when substituting the formulas for the sum of an arithmetic progression.

857 Easy

857 of 998

Application +4 | -1

\frac{1}{p+q}, \frac{1}{q+r}

and

\frac{1}{r+p}

are in

\mathrm{AP}

, then

View answer and explanation

Correct option: B

q^2, p^2

and

r^2

are in AP

Option B (

q^2, p^2, r^2

are in AP) is correct because it is derived from the given AP condition on the reciprocals of (p+q), (q+r), and (r+p).

858 Easy

858 of 998

Understanding +4 | -1

The coefficient of

x^{99}

in the expansion of

(x-1)(x-2) \ldots \ldots(x-100)

, is

View answer and explanation

Correct option: C

-5050

Option C (-5050) is correct because the coefficient of

x^{{99}}

is the negative sum of the roots of the polynomial.

859 Easy

859 of 998

Analyzing +4 | -1

1 + \frac{4}{5} + \frac{7}{{{5^2}}} + \frac{{10}}{{{5^3}}} + \ldots \mathrm{to}\;\infty

View answer and explanation

Correct option: C

\frac{{35}}{{16}}

Option C (

\frac{{35}}{{16}}

) is correct because it is the sum to infinity of the given arithmetico-geometric progression, calculated using the formula

{S_\infty } = \frac{a}{{1 - r}} + \frac{{dr}}{{{{\left( {1 - r} \right)}^2}}}

860 Easy

860 of 998

Application +4 | -1

If arithmetic means between two numbers

a

and

b

are

{A}_1

and

{A}_2,

geometric means are

G_1

and

G_2

, harmonic means are

H_1

and

H_2

, then

\frac{A_1+A_2}{H_1+H_2}=

View answer and explanation

Correct option: C

\frac{{G}_1 {G}_2}{{H}_1 {H}_2}

The correct answer is C (

\frac{{G}_1 {G}_2}{{H}_1 {H}_2}

) because it correctly relates the arithmetic, geometric, and harmonic means between two numbers.

861 Easy

861 of 998

Application +4 | -1

a, b, c

and

d

are four positive real numbers such that

a b c d=1

, the minimum value of

(1+a)(1+b)(1+c)(1+d)

View answer and explanation

Correct option: C

Option B (

3 a_1 a_5

) is correct because the expression

a_1 a_2+a_2 a_3+a_3 a_4+a_4 a_5

simplifies to

4a_1 a_5

when

a_1, a_2, a_3, a_4, a_5

are in Harmonic Progression.

862 Easy

862 of 998

Analyzing +4 | -1

The sum of series

\sum \limits_{n = 1}^\infty \frac{{2n}}{{\left( {2n + 1} \right)!}}

View answer and explanation

Correct option: B

{e^{ - 1}}

Option B (

{e^{ - 1}}

) is correct because the series can be rewritten and simplified to equal

{e^{ - 1}}

863 Easy

863 of 998

Application +4 | -1

a

and

x

are positive integers such that

x < a

and

\sqrt{a-x}, \sqrt{x}, \sqrt{a+x}

are in A.P., then least possible value of

a

View answer and explanation

Correct option: A

Option A (5) is correct because it is the smallest possible integer value for 'a' that satisfies the given conditions for the arithmetic progression.

864 Easy

864 of 998

Analyzing +4 | -1

\frac{x(y+z-x)}{\log x}=\frac{y(z+x-y)}{\log y}=\frac{z(x+y-z)}{\log z}

, then

x^y \,y^x=z^y \,y^z

is equal to

View answer and explanation

Correct option: A

z^x x^z

Option A (

z^x x^z

) is correct because, under the given condition,

x^y \cdot y^x = z^y \cdot y^z = z^x \cdot x^z

865 Easy

865 of 998

Application +4 | -1

The sum of first

n

terms of the series

1^2+2\left(2^2\right)+3^2+2\left(4^2\right)+5^2+2\left(6^2\right)+\ldots

n(n+1)^2 / 2

when

n

is even. When

n

is odd, the sum is

View answer and explanation

Correct option: B

n^2(n+1) / 2

Option C (

\frac{{1}}{{3}}

) is correct. By setting up the equations based on the given information and solving for the common ratio, we find r = 1/3.

866 Easy

866 of 998

Application +4 | -1

{x^{\frac{3}{2}\left( {{{\log }_2}x - 3} \right)}} = \frac{1}{8},

then

x

equals to

View answer and explanation

Correct option: A

Option A (2) is correct because it is the solution to the given exponential equation after applying logarithms and solving the resulting quadratic equation.

867 Easy

867 of 998

Application +4 | -1

\frac{{\log a}}{3} = \frac{{\log b}}{4} = \frac{{\log c}}{5},

then

ca

equals

View answer and explanation

Correct option: B

{b^2}

Option B (

{b^2}

) is correct because it follows directly from the properties of logarithms and exponentials.

868 Easy

868 of 998

Application +4 | -1

{2^x}{.3^{2x}} = 100,

then

x

belongs to

View answer and explanation

Correct option: C

\left( {1,\;2} \right)

Option C is correct because by simplifying the expression and taking the logarithm, we find that x lies between 1 and 2.

869 Easy

869 of 998

Application +4 | -1

{3^{2\sin \,2\alpha - 1}},\,\,\,14\,\,\,\mathrm{and }\,\,{3^{4 - 2\sin \,2\alpha }}

are the first three terms of an

AP

for some

\alpha ,

then the sixth term of this

AP

View answer and explanation

Correct option: D

The correct answer is D (66) because it is the sixth term of the arithmetic progression, calculated using the given condition that the first three terms form an AP.

870 Easy

870 of 998

Analyzing +4 | -1

The sum of the series

1/(1 + {1^2} + {1^4}) + 1/(1 + {2^2} + {2^4}) + 1/(1 + {3^2} + {3^4}) + ...

n

terms is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the series can be expressed as a telescoping series, allowing for simplification and calculation of the sum.

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