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8. sequence and series — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS8. sequence and series1 Mark
MCQ
Let $S_{ k }=\frac{1+2+\ldots .+ K }{ K }$ and $\sum_{j=1}^n S_j^2=\frac{n}{A}\left( Bn ^2+ Cn + D \right)$, where $A , B , C , D \in N$ and $A$ has least value. Then 
  • ✓
    $A + B$ is divisible by $D$
  • B
    $A+B=5(D-C)$
  • C
    $A + C + D$ is not divisible by $B$
  • D
    $A + B + C + D$ is divisible by $5$
✓

Answer

Correct option: A.
$A + B$ is divisible by $D$
a
$S _{ k }=\frac{ k +1}{2}$

$S _{ k }^2=\frac{ k ^2+1+2 k }{4}$

$\therefore \sum \limits_{ j -1}^{ n } S _{ j }^2=\frac{1}{4}\left[\frac{ n ( n +1)(2 n +1)}{6}+ n + n ( n +1)\right]$

$=\frac{ n }{4}\left[\frac{( n +1)(2 n +1)}{6}+1+ n +1\right]$

$=\frac{ n }{4}\left[\frac{2 n ^2+3 n +1}{6}+ n +2\right]$

$=\frac{ n }{4}\left[\frac{2 n ^2+9 n +13}{6}\right]=\frac{ n }{24}\left[2 n ^2+9 n +13\right]$

$A =24, B =2, C =9, D =13$

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