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प्रश्न
Find the sum to n terms of the series whose nth terms is given by (2n – 1)2
उत्तर
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संबंधित प्रश्न
Find the sum to n terms of the series `1/(1xx2) + 1/(2xx3)+1/(3xx4)+ ...`
Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …
Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
13 + 33 + 53 + 73 + ...
22 + 42 + 62 + 82 + ...
1.2.5 + 2.3.6 + 3.4.7 + ...
1.2.4 + 2.3.7 +3.4.10 + ...
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...
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1 + 3 + 6 + 10 + 15 + ...
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . .\]
\[\frac{1}{1 . 6} + \frac{1}{6 . 11} + \frac{1}{11 . 14} + \frac{1}{14 . 19} + . . . + \frac{1}{(5n - 4) (5n + 1)}\]
The value of \[\sum^n_{r = 1} \left\{ (2r - 1) a + \frac{1}{b^r} \right\}\] is equal to
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If Sn = \[\sum^n_{r = 1} \frac{1 + 2 + 2^2 + . . . \text { Sum to r terms }}{2^r}\], then Sn is equal to
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Write the 50th term of the series 2 + 3 + 6 + 11 + 18 + ...
If \[1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .\] to n terms is S, then S is equal to
Sum of n terms of the series \[\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} +\] ....... is
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Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + … to n terms
Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then `sum_(r = 1)^n S_r/s_r` equals ______.
The sum of the series `1/(x + 1) + 2/(x^2 + 1) + 2^2/(x^4 + 1) + ...... + 2^100/(x^(2^100) + 1)` when x = 2 is ______.
The sum of all natural numbers 'n' such that 100 < n < 200 and H.C.F. (91, n) > 1 is ______.
