Let a _ n be a sequence such that a_1+a_2+ +a_n=n^2+3 n (n+1)(n+2… - Vidyadip

MCQ

Let $\left\langle a _{ n }\right\rangle$ be a sequence such that $a_1+a_2+\ldots+a_n=\frac{n^2+3 n}{(n+1)(n+2)}$. If $28 \sum \limits_{ k =1}^{10} \frac{1}{ a _{ k }}= p _1 p _2 p _3 \ldots p _{ m }$, where $p _1, p _2, \ldots . pm$ are the first $m$ prime numbers, then $m$ is equal to

A
$7$
✓
$6$
C
$5$
D
$8$

✓

Answer

Correct option: B.

$6$

b
$a_n=S_n-S_{n-1}=\frac{ n ^2+3 n }{( n +1)( n +2)}-\frac{( n -1)( n +2)}{ n ( n +1)}$

$\Rightarrow a _{ n }=\frac{4}{ n ( n +1)( n +2)}$

$\Rightarrow 28 \sum \limits_{ k =1}^{10} \frac{1}{ a _{ k }}=28 \sum \limits_{ k =1}^{10} \frac{ k ( k +1)( k +2)}{4}$

$=\frac{7}{4} \sum \limits_{ k =1}^{10}( k ( k +1)( k +2)( k +3)-( k -1) k ( k +1)( k +2))$

$=\frac{7}{4} \cdot 10 \cdot 11 \cdot 12 \cdot 13=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$

So $m =6$

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