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Find the sum of the series whose nth term is given by:
(3n2 + 2n)
n (n + 1) (n + 4)
(4n3 + 6n2 + 2n)
(3n2 – 3n + 2)
(2n2 – 3n + 5)
(n3 – 3n)
Find the sum of the series:
(22 + 42 + 62 + 82 + … to n terms)
(23 + 43 + 63 + 83 + … to n terms)
(52 + 62 + 72 + … + 202)
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + … to n terms
(3 × 8) + (6 × 11) + (9 × 14) + … to n terms
(1 × 22) + (2 × 32) + (3 × 42) + … to n terms
(1 × 22) + (3 × 32) + (5 × 42) + … to n terms
(3 × 12) + (5 × 22) + (7 × 32) + … to n terms
(1 × 2 × 3) + (2 × 3 × 4) + (3 × 4 × 5) + … to n terms
(1 × 2 × 4) + (2 × 3 × 7) + (3 × 4 × 10) + … to n terms
…. To n terms
to n terms
3 + 15 + 35 + 63 +...to n terms
1 + 5 + 12 + 22 + 35 +... to n terms
5 + 7 + 13 + 31 + 85 + …. To n terms
If, prove that,
.
If Sn denotes the sum of the cubes of the first n natural numbers and sn denotes the sum of the first n natural numbers then find the value of .
Find the sum (2 + 4 + 6 + 8 +… + 100).
Find the sum (41 + 42 + 43 + …. + 100).
Find the sum 112 + 122 + 132+ …202
Find the sum 63 + 73 + 83 + 93 + 103.
If , find the value of .
If, find the value of.
Find the sum of the series {22 + 42 + 62 + …. + (2n)2}
Find the sum of 10 terms of the geometric series
Find the sum of n terms of the series whose rth term is (r + 2r).