For a positive integer n- let a n -1-1-2-1-3-1-4-1-2 n -1 thenA- a 100-100B- a 100-100C- a 200-100D- a 100-100

The correct option is B a(100) lt; 100 a(n) = 1 + (1/2+1/3 )+ ( 1/4+1/5+1/6+1/7 )+ (1/8+1/9+…+1/15 )…+ ( 1/2^n 1+…+ 1/2^n 1 ) ⇒ a(n) lt;1+ ( 1/2+1/2 )+ ( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of n Terms

For a positiv...

Question

For a positive integer $n,$ let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \dots + \frac{1}{2^{n} - 1}$ then

$a (100) < 100$

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

$a (100) > 100$

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

$a (200) < 100$

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

$a (100) = 100$

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

Open in App

Solution

The correct option is A
$a (100) < 100$

$a (n) = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}) + (\frac{1}{8} + \frac{1}{9} + \dots + \frac{1}{15}) \dots + (\frac{1}{2^{n - 1}} + \dots + \frac{1}{2^{n} - 1})$

$\Rightarrow a (n) < 1 + (\frac{1}{2} + \frac{1}{2}) + (\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4}) + \dots$

$\Rightarrow a (n) < 1 + (n - 1)$

$\Rightarrow a (n) < n$

$\Rightarrow a (100) < 100$

Suggest Corrections

Similar questions

Q. If a₁ = 1 and a_n+1 – 3a_n + 2 = 4n for every positive integer n, then a₁₀₀ equals

Q. If

a_{1} = 1

and

a_{n + 1} - 3 a_{n} + 2 = 4 n

for every positive integer n, then

a_{100}

equals

Q. If a₁ = 1 and a_n+1 – 3a_n + 2 = 4n for every positive integer n, then a₁₀₀ equals (CAT 2005)

Q. Let

A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & \frac{- 1 - i \sqrt{3}}{2} \frac{- 1 + i \sqrt{3}}{2} & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

. Then,

A^{100} =

Q. If

a_{1} = 1

and

a_{n + 1} - 3 a_{n} + 2 = 4 n

for every positive integer n, then

a_{100}

equals (CAT 2005)

Join BYJU'S Learning Program

For a positive integer n, let a(n)=1+12+13+14…+12n−1 then

The correct option is A a(100)<100 a(n)=1+(12+13)+(14+15+16+17)+(18+19+…+115)…+(12n−1+…+12n−1) ⇒a(n)<1+(12+12)+(14+14+14+14)+… ⇒a(n)<1+(n−1) ⇒a(n)<n ⇒a(100)<100

For a positive integer $n,$ let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \dots + \frac{1}{2^{n} - 1}$ then