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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2021 · 31 August · Morning Shift · Mathematics · Sequences and Series

The sum of 10 terms of the series 3 1 ^ 2 × 2 ^ 2 + 5 2 ^ 2 × 3 ^ 2...

The sum of 1010 terms of the series 312×22+522×32+732×42+…\dfrac{3}{{1}^{2} \times {2}^{2}} + \dfrac{5}{{2}^{2} \times {3}^{2}} + \dfrac{7}{{3}^{2} \times {4}^{2}} + \ldots is :

Options

  1. A143144\dfrac{143}{144}
  2. B99100\dfrac{99}{100}
  3. C11
  4. D120121\dfrac{120}{121}

Correct answer

D. 120121\dfrac{120}{121}

Step-by-step solution

Given that

312×22+522×32+732×42+…\dfrac{3}{{1}^{2} \times {2}^{2}} + \dfrac{5}{{2}^{2} \times {3}^{2}} + \dfrac{7}{{3}^{2} \times {4}^{2}} + \ldots

Here 3,5,7,−−−−∈A.P3, 5, 7, - - - - \in A . P

First term a=3 and a = 3 \text{ and }Common difference d=2,tr=a+(r−1)d=3+(r−1)2=2r+1d = 2, {t}_{r} = a + \left(r - 1\right)d = 3 + \left(r - 1\right)2 = 2r + 1

=∑r=110(2r+1)r2⋅(r+1)2= \mathop{∑}_{r = 1}^{10}\dfrac{( 2r + 1 )}{{r}^{2} \cdot ( r + 1{)}^{2}}

=∑r=110r2+2r+1−r2r2⋅(r+1)2= \mathop{∑}_{r = 1}^{10}\dfrac{{r}^{2} + 2r + 1 - {r}^{2}}{{r}^{2} \cdot ( r + 1{)}^{2}}

=∑r=110(r+1)2−r2r2⋅(r+1)2= \mathop{∑}_{r = 1}^{10}\dfrac{( r + 1{)}^{2} - {r}^{2}}{{r}^{2} \cdot ( r + 1{)}^{2}}

=∑r=110{1r2−1(r+1)2}= \mathop{∑}_{r = 1}^{10}\left\{\dfrac{1}{{r}^{2}} - \dfrac{1}{( r + 1{)}^{2}}\right\}

=(112−122)+(122−132)+(132−142)+…..+(1102−1112)=1−1121=120121= \left(\dfrac{1}{{1}^{2}} - \dfrac{1}{{2}^{2}}\right) + \left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}\right) + \left(\dfrac{1}{{3}^{2}} - \dfrac{1}{{4}^{2}}\right) + \ldots . . + \left(\dfrac{1}{{10}^{2}} - \dfrac{1}{{11}^{2}}\right) = 1 - \dfrac{1}{121} = \dfrac{120}{121}

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