Sum of the series $1 + 2.2 + {3.2}^{2} + {4.2}^{3} + . . + {100.2}^{99}$ is

{100.2}^{100} + 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{99.2}^{100} + 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{99.2}^{100} - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{100.2}^{100} - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C ${99.2}^{100} + 1$
It is been clear that the series is in Arithmetic-Geometric progression with
$A . P = 1 + 2 + 3 + . . . . . . . + 100 a n d, G . P = 1 + 2 + 2^{2} + . . . . . + 2^{99} L e t, S = 1 + 2.2 + 3. 2^{2} + 4. 2^{3} + . . . . . + 100. 2^{99} ⟶ (1) Multiply the above eauation by 2 2 S = 2 + 2. 2^{2} + 3. 2^{3} + . . . . . + 99. 2^{99} + 100. 2^{100} ⟶ (2) n o w s u b t r a c t (2) f r o m (1) w e g e t, S - 2 S = (1 + 2 + 2^{2} + . . . . + 2^{99}) - 100. 2^{100} \Rightarrow - S = \frac{(2^{100} - 1)}{2 - 1} - 100. 2^{100} (s i n c e t h e a b o v e s e r i e s i n b r a c k e t i s i n G . P s o S_{n} = \frac{a (r^{n} - 1)}{r - 1} a n d t h i s i s a G . P w i t h 100 t e r m s a n d a = 1; r = 2) \Rightarrow S = - 2^{100} + 1 + 100. 2^{100} ∴ S = 99. 2^{100} + 1$

Suggest Corrections

Similar questions

1 + 2.2 + {3.2}^{2} + {4.2}^{3} + {5.2}^{4} + \dots + {100.2}^{99}

1 + 2.2 + 3. 2^{2} + 4. 2^{4} + 5. 2^{4} + . . . + 100. 2^{99}

Find the sum of the series $1.2 + {2.2}^{2} + {3.2}^{2} + . . . . . .100 {.2}^{100}$

1 + 2.2 + {3.2}^{2} + . . . . + 100 \times 2^{99}

Join BYJU'S Learning Program