Prove that the following equations hold for any natural n- 1-1-2 - 1-2-3 - 1-3-4 - - 1-n n - 1 - n-n - 1-

a_n=1n(n+1)=1n 1n+1 a_1=11 12 a_2=12 13 a_3=13 14 a_4=14 15 . . a_n=1n(n+1)=1n 1n+1 Adding all a_1+a_2+...+a_n=11 1 ...

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

Byju's Answer

Standard X

Mathematics

Sum of N Terms of an AP

Prove that th...

Question

Prove that the following equations hold for any natural n.
$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . \frac{1}{n (n + 1)} = \frac{n}{n + 1} .$

Open in App

Solution

$a_{n} = \frac{1}{n (n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$

$a_{1} = \frac{1}{1} - \frac{1}{2}$

$a_{2} = \frac{1}{2} - \frac{1}{3}$

$a_{3} = \frac{1}{3} - \frac{1}{4}$

$a_{4} = \frac{1}{4} - \frac{1}{5}$
$.$
$.$
$a_{n} = \frac{1}{n (n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$

Adding all-

$a_{1} + a_{2} + . . . + a_{n} = \frac{1}{1} - \frac{1}{n + 1}$

$= \frac{n}{n + 1}$

Suggest Corrections

Similar questions

Q. Prove that

\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . . + \frac{1}{n (n + 1)} = \frac{n}{n + 1}

Prove by using the principle of mathematical induction that $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . + \frac{1}{n (n + 1)} = \frac{n}{n + 1}$

$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . . . . . + \frac{1}{n (n + 1)} = \frac{n}{n + 1}$

Q. Find m if the following equation holds true

1.2 + 2.3 + 3.4 + . . + n (n + 1) = \frac{1}{m} n (n + 1) (n + 2)

Q. Check If the equation is true or false .

\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)} = \frac{n}{n + 1}

Join BYJU'S Learning Program

Prove that the following equations hold for any natural n.11.2+12.3+13.4+...1n(n+1)=nn+1.

an=1n(n+1)=1n−1n+1a1=11−12a2=12−13a3=13−14a4=14−15..an=1n(n+1)=1n−1n+1Adding all-a1+a2+...+an=11−1n+1=nn+1

Prove that the following equations hold for any natural n.
$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . \frac{1}{n (n + 1)} = \frac{n}{n + 1} .$