Advertisements
Advertisements
प्रश्न
\[\frac{1}{1 . 6} + \frac{1}{6 . 11} + \frac{1}{11 . 14} + \frac{1}{14 . 19} + . . . + \frac{1}{(5n - 4) (5n + 1)}\]
उत्तर
Let \[T_n\] be the nth term of the given series.
Thus, we have: \[T_n = \frac{1}{(5n - 4) (5n + 1)}\]
Now, let \[S_n\] be the sum of n terms of the given series.
Thus, we have:
\[S_n = \sum^n_{k = 1} \frac{1}{\left( 5k - 4 \right)\left( 5k + 1 \right)}\]
\[ = \frac{1}{5} \sum^n_{k = 1} \left( \frac{1}{\left( 5k - 4 \right)} - \frac{1}{\left( 5k + 1 \right)} \right)\]
\[ = \frac{1}{5} \sum^n_{k = 1} \frac{1}{\left( 5k - 4 \right)} - \frac{1}{5} \sum^n_{k = 1} \frac{1}{\left( 5k + 1 \right)}\]
\[ = \frac{1}{5}\left[ \left( 1 + \frac{1}{6} + \frac{1}{11} + \frac{1}{16} + . . . + \frac{1}{5n - 4} \right) - \left( \frac{1}{6} + \frac{1}{11} + \frac{1}{16} + . . . + \frac{1}{5n - 4} + \frac{1}{5n + 1} \right) \right]\]
\[ = \frac{1}{5}\left[ 1 - \left( \frac{1}{5n + 1} \right) \right]\]
\[ = \frac{n}{5n + 1}\]
APPEARS IN
संबंधित प्रश्न
Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …
Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
Find the sum to n terms of the series whose nth terms is given by n2 + 2n
22 + 42 + 62 + 82 + ...
1.2.4 + 2.3.7 +3.4.10 + ...
3 × 12 + 5 ×22 + 7 × 32 + ...
Find the sum of the series whose nth term is:
2n3 + 3n2 − 1
Find the sum of the series whose nth term is:
n (n + 1) (n + 4)
Write the sum of the series 12 − 22 + 32 − 42 + 52 − 62 + ... + (2n − 1)2 − (2n)2.
1 + 3 + 7 + 13 + 21 + ...
3 + 7 + 14 + 24 + 37 + ...
1 + 4 + 13 + 40 + 121 + ...
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . .\]
The value of \[\sum^n_{r = 1} \left\{ (2r - 1) a + \frac{1}{b^r} \right\}\] is equal to
If Sn = \[\sum^n_{r = 1} \frac{1 + 2 + 2^2 + . . . \text { Sum to r terms }}{2^r}\], then Sn is equal to
Write the sum of 20 terms of the series \[1 + \frac{1}{2}(1 + 2) + \frac{1}{3}(1 + 2 + 3) + . . . .\]
Write the 50th term of the series 2 + 3 + 6 + 11 + 18 + ...
Let Sn denote the sum of the cubes of first n natural numbers and sn denote the sum of first n natural numbers. Then, write the value of \[\sum^n_{r = 1} \frac{S_r}{s_r}\] .
The sum to n terms of the series \[\frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + . . . . + . . . .\] is
The sum of the series
\[\frac{1}{\log_2 4} + \frac{1}{\log_4 4} + \frac{1}{\log_8 4} + . . . . + \frac{1}{\log_2^n 4}\] is
Sum of n terms of the series \[\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} +\] ....... is
The sum of the series \[\frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{80}{81} +\] to n terms is
If \[\sum^n_{r = 1} r = 55, \text{ find } \sum^n_{r = 1} r^3\] .
3 + 5 + 9 + 15 + 23 + ...
2 + 5 + 10 + 17 + 26 + ...
Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + … to n terms
Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + ... to 10 terms
The sum of all natural numbers 'n' such that 100 < n < 200 and H.C.F. (91, n) > 1 is ______.
The sum `sum_(k = 1)^20k 1/2^k` is equal to ______.
