The correct option is C n+12 We have, (1+12)(1+13)(1+14)..............(1+1n) ⇒(32)(43)(54)..............(n+1n) All terms are cancel out, only ...

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

Summation by Sigma Method

1+121+131+14....

Question

$(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) . . . . . . . . . . . . . . (1 + \frac{1}{n}) =$ :

n

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

\frac{n - 1}{2}

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

\frac{n + 1}{2}

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

\frac{n}{2}

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

Open in App

Solution

The correct option is C $\frac{n + 1}{2}$
We have,
$(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) . . . . . . . . . . . . . . (1 + \frac{1}{n})$
$\Rightarrow (\frac{3}{2}) (\frac{4}{3}) (\frac{5}{4}) . . . . . . . . . . . . . . (\frac{n + 1}{n})$

All terms are cancel out, only $2$ will be left.

So,
$\Rightarrow \frac{n + 1}{2}$

Hence, this is the answer.

Suggest Corrections

Similar questions

\frac{3}{1 \cdot 2} (\frac{1}{2}) + \frac{4}{2 \cdot 3} {(\frac{1}{2})}^{2} + \frac{5}{3 \cdot 4} {(\frac{1}{2})}^{3} + \dots

n

1 + 2 + 3 + 4 + . . . . . + n = ?

1 + 2 + 3 + 4 + . . . . . + (n - 1) = ?

1 + 1 + 1 + 1 + . . . . . + n = ?

1 + 1 + 1 + 1 + . . . . . . + (n - 1) = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + {(n - 1)}^{2} = ?

1 \times n + 2 (n - 1) + 3 (n - 2) + \dots \dots + (n - 1) \times 2 + n \times 1

\frac{{2.3}^{n + 1} + {7.3}^{n - 1}}{3^{n + 2} - 2 {(\frac{1}{3})}^{1 - n}} =

\frac{1}{n!} + \frac{1}{2! (n - 2)!} + \frac{1}{4! (n - 4)!} + = \frac{1}{n!} 2^{n - 1}

Join BYJU'S Learning Program