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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
$\sum\limits_{r = 0}^{100} {({r^2} + 4r + 4)\left| \!{\underline {\,
  {r + 1} \,}} \right. } $ is equal to :-
  • ✓
    $\left| \!{\underline {\,
      {103} \,}} \right.  - 2$
  • B
    $\left| \!{\underline {\,
      {102} \,}} \right.  - 1$
  • C
    ${\left( {\left| \!{\underline {\,
      {102} \,}} \right. } \right)^2}$
  • D
    None
✓

Answer

Correct option: A.
$\left| \!{\underline {\,
  {103} \,}} \right.  - 2$
a
$\sum\limits_{r = 0}^{100} {{{\left( {r + 2} \right)}^2}r + 1!}  = \sum\limits_{r = 0}^{100} {\left( {r + 2} \right)} \,r + 2!$

$\sum\limits_{r = 0}^{100} {\left( {r + 3 - 1} \right)} \,r + 2! = \sum\limits_{r = 0}^{100} {r + 3!}  - \sum\limits_{r = 0}^{100} {r + 2!} $

$ \Rightarrow \left( {3! + 4! + 5! + ... + 103!} \right) - \left( {2! + 3! + 4! + ... + 102!} \right)$

$ \Rightarrow 103! - 2$

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