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JEE Main 20236 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2023 · 6 April · Morning Shift · Mathematics · Sequences and Series

The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 +...

The sum of the first 2020 terms of the series 5+11+19+29+41+...5 + 11 + 19 + 29 + 41 + . . . is

Options

  1. A35203520
  2. B34503450
  3. C32503250
  4. D34203420

Correct answer

A. 35203520

Step-by-step solution

Let the given series be Sn{S}_{n}, then we can write as

Sn=5+11+19+29+41+............+TnSn=5+11+19+29+41+......+Tn−1+Tn{S}_{n} = 5 + 11 + 19 + 29 + 41 + . . . . . . . . . . . . + {T}_{n}{S}_{n} = 5 + 11 + 19 + 29 + 41 + . . . . . . + {T}_{n - 1} + {T}_{n}

Subtracting above equations, we get

0=5+6+8+10+....−Tn0 = 5 + 6 + 8 + 10 + . . . . - {T}_{n}

⇒Tn=5+[(n−1)2(12+(n−2)2)]⇒ {T}_{n} = 5 + \left[\dfrac{\left(n - 1\right)}{2}\left(12 + \left(n - 2\right)2\right)\right]

⇒Tn=5+(n−1)(n+4)⇒ {T}_{n} = 5 + \left(n - 1\right)\left(n + 4\right)

⇒Tn=n2+3n+1⇒ {T}_{n} = {n}^{2} + 3n + 1

So,

S20=∑n=120Tn{S}_{20} = \mathop{∑}_{n = 1}^{20}{T}_{n}

⇒S20=∑n=120(n2+3n+1)⇒ {S}_{20} = \mathop{∑}_{n = 1}^{20}\left({n}^{2} + 3n + 1\right)

⇒S20=20×21×416+3×20×212+20⇒ {S}_{20} = \dfrac{20 \times 21 \times 41}{6} + \dfrac{3 \times 20 \times 21}{2} + 20

⇒S20=3520⇒ {S}_{20} = 3520

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