Advertisements
Advertisements
Question
Write the sum of the series 2 + 4 + 6 + 8 + ... + 2n.
Solution
\[S_n = 2 + 4 + 6 + 8 + . . . + 2n\]
\[= \frac{n}{2}\left[ 4 + \left( n - 1 \right)2 \right]\]
\[ = \frac{n}{2}\left[ 2 + 2n \right]\]
\[ = n\left[ n + 1 \right]\]
APPEARS IN
RELATED QUESTIONS
Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
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 12 + (12 + 22) + (12 + 22 + 32) + …
Find the sum to n terms of the series whose nth terms is given by (2n – 1)2
Show that `(1xx2^2 + 2xx3^2 + ...+nxx(n+1)^2)/(1^2 xx 2 + 2^2 xx3 + ... + n^2xx (n+1))` = `(3n + 5)/(3n + 1)`
13 + 33 + 53 + 73 + ...
22 + 42 + 62 + 82 + ...
1.2.5 + 2.3.6 + 3.4.7 + ...
1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...
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:
n3 − 3n
Find the sum of the series whose nth term is:
n (n + 1) (n + 4)
Find the 20th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 + ...
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 + ...
\[\frac{1}{1 . 6} + \frac{1}{6 . 11} + \frac{1}{11 . 14} + \frac{1}{14 . 19} + . . . + \frac{1}{(5n - 4) (5n + 1)}\]
The value of \[\sum^n_{r = 1} \left\{ (2r - 1) a + \frac{1}{b^r} \right\}\] is equal to
If ∑ n = 210, then ∑ n2 =
If Sn = \[\sum^n_{r = 1} \frac{1 + 2 + 2^2 + . . . \text { Sum to r terms }}{2^r}\], then Sn is equal to
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 of the series 12 + 32 + 52 + ... to n terms is
The sum of the series \[\frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{80}{81} +\] to n terms is
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
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 ______.
If |x| < 1, |y| < 1 and x ≠ y, then the sum to infinity of the following series:
(x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + .... is ______.
The sum `sum_(k = 1)^20k 1/2^k` is equal to ______.
