Prove that 11-2-12-3-13-4-1nn-1-nn-1-

Now, 11.2+12.3+13.4+.....+1n(n+1) =(1 12)+(12 13)+.....+(1n 1n+1) =1 1n+1 [ Cancelling the similar terms] =nn+1 . ...

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

Byju's Answer

Standard V

Mathematics

Multiplication of Fractions

Prove that ...

Question

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

Open in App

Solution

Now,
$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . . + \frac{1}{n (n + 1)}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + . . . . . + (\frac{1}{n} - \frac{1}{n + 1})$
$= 1 - \frac{1}{n + 1}$ [ Cancelling the similar terms]
$= \frac{n}{n + 1}$ .

Suggest Corrections

Similar questions

Q. 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} .

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. 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}

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

equals
[AMU 1995; RPET 1996; UPSEAT 1999, 2001]

Join BYJU'S Learning Program

Prove that 11.2+12.3+13.4+.....+1n(n+1)=nn+1.

Now,11.2+12.3+13.4+.....+1n(n+1)=(1−12)+(12−13)+.....+(1n−1n+1)=1−1n+1 [ Cancelling the similar terms]=nn+1.

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

Now,
$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . . + \frac{1}{n (n + 1)}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + . . . . . + (\frac{1}{n} - \frac{1}{n + 1})$
$= 1 - \frac{1}{n + 1}$ [ Cancelling the similar terms]
$= \frac{n}{n + 1}$ .