A sequence is satisfying a1+a2+a3+......+an=n2an∀n∈N and if a1=23 then which of the following is/ are correct?

a1+a2+a3+...+an−1+an=n2an ----- (1)Replacing n by n – 1 a1+a2+....+an−2+an−1=(n−1)2an−1 …. (2)(1) – (2) ⇒an=n2.an−(n−1)2an−1⇒(n2−1)an=(n−1)2an−1 ⇒(n+1)an=(n−1)am−1 ⇒anan−1=n−1n+1 a2a1a3a2a4a3....an−1an−2.anan−1=13243546....n−2nn−1n+1 ⇒ana1=1.2n(n+1)

A sequence is satisfying a1+a2+a3+......+an=n2an∀n∈N and if a1=23 then which of the following is/ are correct?

a1+a2+a3+...+an−1+an=n2an ----- (1)Replacing n by n – 1 a1+a2+....+an−2+an−1=(n−1)2an−1 …. (2)(1) – (2) ⇒an=n2.an−(n−1)2an−1⇒(n2−1)an=(n−1)2an−1 ⇒(n+1)an=(n−1)am−1 ⇒anan−1=n−1n+1 a2a1a3a2a4a3....an−1an−2.anan−1=13243546....n−2nn−1n+1 ⇒ana1=1.2n(n+1)

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A sequence $< a_{n} >$ is satisfying $a_{1} + a_{2} + a_{3} + ...... + a_{n} = n^{2} a_{n} \forall n \in N$ and if $a_{1} = 23$ then which of the following is/ are correct?

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$a_{100} = \frac{23}{5050}$

Integral part of $\frac{a_{46}}{a_{23}} = 0$

$a_{50} < a_{49}$

Fractional part of $a_{20}$ is $\frac{1}{3}$

answer is A, B, C.

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Detailed Solution

$a_{1} + a_{2} + a_{3} + ... + a_{n - 1} + a_{n} = n^{2} a_{n}$ ----- (1)
Replacing n by n – 1
$a_{1} + a_{2} + .... + a_{n - 2} + a_{n - 1} = {(n - 1)}^{2} a_{n - 1}$ …. (2)
(1) – (2) $\Rightarrow a_{n} = n^{2} . a_{n} - {(n - 1)}^{2} a_{n - 1}$

$\Rightarrow (n^{2} - 1) a_{n} = {(n - 1)}^{2} a_{n - 1} \Rightarrow (n + 1) a_{n} = (n - 1) a_{m - 1} \Rightarrow \frac{a_{n}}{a_{n - 1}} = \frac{n - 1}{n + 1} \frac{a_{2}}{a_{1}} \frac{a_{3}}{a_{2}} \frac{a_{4}}{a_{3}} .... \frac{a_{n - 1}}{a_{n - 2}} . \frac{a_{n}}{a_{n - 1}} = \frac{1}{3} \frac{2}{4} \frac{3}{5} \frac{4}{6} .... \frac{n - 2}{n} \frac{n - 1}{n + 1} \Rightarrow \frac{a_{n}}{a_{1}} = \frac{1.2}{n (n + 1)}$