If \frac11^4 + \frac12^4 + \frac13^4 + \ldots \infty= \frac\pi^490 , \frac11^4 + \frac13^4 + \frac15^4 + \ldots \infty= \alpha , \frac12^4 + \frac14^4 + \frac16^4 + \ldots \infty= \beta , then \frac\alpha\beta is equal to :

Sign in

Sequences and Series question

2025 · 8 Apr · Shift 2 · Q28

Sequences and Series question

2025 · 8 Apr · Shift 2 · Q28

View written solutionFree

Correct answer: D

Let

S=\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}.

We are given:

odd terms sum

\alpha=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\cdots

even terms sum

\beta=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\cdots

Clearly,

S=\alpha+\beta.

Now compute the even terms sum $\beta$ :

\beta=\sum_{n=1}^{\infty}\frac{1}{(2n)^4} =\sum_{n=1}^{\infty}\frac{1}{16n^4} =\frac{1}{16}\sum_{n=1}^{\infty}\frac{1}{n^4} =\frac{1}{16}S.

So,

\beta=\frac{1}{16}S.

Therefore,

\alpha=S-\beta=S-\frac{1}{16}S=\frac{15}{16}S.

Now find the ratio:

\frac{\alpha}{\beta}=\frac{\frac{15}{16}S}{\frac{1}{16}S}=15.

Hence the correct option is

\boxed{15}

which is option D.

More from Sequences and Series

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , then $a_{6}$ is equal to:2025 · MCQ
Suppose that the number of terms in an A.P. is $2 k, k \in N$ . If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to:2025 · MCQ
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to2025 · MCQ
The roots of the quadratic equation $3 x^{2} - p x + q = 0$ are $1 0^{th}$ and $1 1^{th}$ terms of an arithmetic progression with common difference $\frac{3}{2}$ . If the sum of the first 11 terms of this arithmetic progression is…2025 · Numerical
Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots$ upto $n$ terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is $2026 S_{2025}$ , then the absolute difference…2025 · MCQ
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to :2025 · MCQ
If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7 ^{2}} (5 + 2 α) + \frac{1}{7 ^{3}} (5 + 3 α) + \dots\dots\dots\dots \infty$ , then the value of $α$ is :2025 · MCQ
Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, \dots$ . Then $k = 1 \sum 100 a_{k}$ is equal to2025 · MCQ

Sign in

Sequences and Series question

2025 · 8 Apr · Shift 2 · Q28

Sequences and Series question

2025 · 8 Apr · Shift 2 · Q28

View written solutionFree

Correct answer: D

Let

S=\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}.

We are given:

odd terms sum

\alpha=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\cdots

even terms sum

\beta=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\cdots

Clearly,

S=\alpha+\beta.

Now compute the even terms sum $\beta$ :

\beta=\sum_{n=1}^{\infty}\frac{1}{(2n)^4} =\sum_{n=1}^{\infty}\frac{1}{16n^4} =\frac{1}{16}\sum_{n=1}^{\infty}\frac{1}{n^4} =\frac{1}{16}S.

So,

\beta=\frac{1}{16}S.

Therefore,

\alpha=S-\beta=S-\frac{1}{16}S=\frac{15}{16}S.

Now find the ratio:

\frac{\alpha}{\beta}=\frac{\frac{15}{16}S}{\frac{1}{16}S}=15.

Hence the correct option is

\boxed{15}

which is option D.