Sum of the series -r-1n1-a r-ba r-a-b- where -0 is

The correct option is A n(a+b)[a(n+1)+b]1(ar+b)(ar+a+b)=1a[1ar+b 1ar+a+b] ∑_r=1^n1(ar+b)(ar+a+b)= ∑_r=1^n1a[1ar+b 1ar+a+b] =1a[1a+b 1a+a+b] +1a[12a+b 12a+a+b ...

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

Byju's Answer

Standard XII

Mathematics

Method of Difference

Sum of the se...

Question

Sum of the series $n \sum r = 1 \frac{1}{(a r + b) (a r + a + b)}$ , (where $a \neq 0$ ) is

\frac{n}{(a + b) [a (n + 1) + b]}

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

\frac{n}{(a + b) [a n + b]}

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

\frac{n}{(a + b) [a (n - 1) + b]}

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

\frac{n}{(a + b) [a (n + 2) + b]}

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

Open in App

Solution

The correct option is A $\frac{n}{(a + b) [a (n + 1) + b]}$
$\frac{1}{(a r + b) (a r + a + b)} = \frac{1}{a} [\frac{1}{a r + b} - \frac{1}{a r + a + b}]$
$n \sum r = 1 \frac{1}{(a r + b) (a r + a + b)} = n \sum r = 1 \frac{1}{a} [\frac{1}{a r + b} - \frac{1}{a r + a + b}] = \frac{1}{a} [\frac{1}{a + b} - \frac{1}{a + a + b}] + \frac{1}{a} [\frac{1}{2 a + b} - \frac{1}{2 a + a + b}] + \frac{1}{a} [\frac{1}{3 a + b} - \frac{1}{3 a + a + b}] + ⋮ ⋮ + \frac{1}{a} [\frac{1}{a n + b} - \frac{1}{a n + a + b}] S_{n} = \frac{1}{a} [\frac{1}{a + b} - \frac{1}{a (n + 1) + b}] = \frac{n}{(a + b) [a (n + 1) + b]}$

Suggest Corrections

Similar questions

Q. Sum of the series

n \sum r = 1 \frac{1}{(a r + b) (a r + a + b)}

, (where

a \neq 0

) is

Q. Let the sum of series be

n \sum r = 1 (\frac{1}{4 r^{2} - 1}) .

\frac{n}{k n + m} .

.Find

k + m

Q. Sum of the series

\sum_{k = 0}^{r} {(- 1)}^{r} (\begin{matrix} n r \end{matrix})

Q. Sum the following series to n terms is

n \sum r = 1 \frac{1}{4 r^{2} - 1}

S_{n} = \frac{1}{k} [1 - \frac{1}{k n + 1}]

S_{\infty} = \frac{1}{k}

. Find

k

Q. Two lines

¯ a z + a ¯ z + c = 0, ¯ b z + b ¯ z + d = 0

where

a, b \in C - {0}

and

c, d \in R

, are

Join BYJU'S Learning Program

Explore more

Method of Difference

Standard XII Mathematics

Join BYJU'S Learning Program

Sum of the series n∑r=11(ar+b)(ar+a+b), (where a≠0) is

The correct option is A n(a+b)[a(n+1)+b]1(ar+b)(ar+a+b)=1a[1ar+b−1ar+a+b] n∑r=11(ar+b)(ar+a+b)=n∑r=11a[1ar+b−1ar+a+b]=1a[1a+b−1a+a+b]+1a[12a+b−12a+a+b]+1a[13a+b−13a+a+b]+ ⋮ ⋮+1a[1an+b−1an+a+b]Sn=1a[1a+b−1a(n+1)+b]=n(a+b)[a(n+1)+b]

Sum of the series $n \sum r = 1 \frac{1}{(a r + b) (a r + a + b)}$ , (where $a \neq 0$ ) is