1 2 2 3 3 4 series A geometric series is a sequence of numbers in which the ratio between any two consecutive terms is always the same and often written in the form a ar ar 2 ar 3 where a is the first term of the series and r is the common ratio 1 r 1
N Let S n 1 2 3 4 cdots n displaystyle sum k 1 n k S n 1 2 3 4 n k 1 n k The elementary trick for solving this equation which Gauss is supposed to have used as a child is a rearrangement of the sum as follows The infinite series whose terms are the natural numbers 1 2 3 4 is a divergent series The n th partial sum of the series is the triangular number which increases without bound as n goes to infinity Because the sequence of partial sums fails to converge to a finite limit the series does not have a sum
1 2 2 3 3 4 series
1 2 2 3 3 4 series
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2 1 2 1 3 8 3 1 2 3 6 1 1 3
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1 2 2 3 4T65E 24203536
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GeeksforGeeks Find n th term in series 1 2 2 3 3 3 4 4 4 4 Given series 1 2 2 3 3 3 4 4 4 4 find n th term of the series The pattern is obvious There is one 1 two 2 s three 3 s etc Examples Input n 5 Output 3 Input n 7 Output 4 Given a number N the task is to find the sum of the below series till N terms Examples Input N 6 Output 0 240476 Input N 10 Output 0 263456 Unmute Approach From the given series find the formula for Nth term 1st term 1 2 2nd term 2 3 3rd term 3 4 4th term 4 5 Nthe term
Explanation 1 2 2 3 2 4 2 6 12 20 Simple Solution One by one add elements recursively Below is the implementation Python3 include using namespace std int sum int n if n 1 return 2 else return n n 1 sum n 1 int main int n 2 cout Answer link Is divergent 1 n n n 1 and sum k 1 oo 1 k sum k 1 oo k k 1 but sum k 1 oo 1 k is the so called harmonic series which is divergent So sum k 1 oo k k 1 is also divergent
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Find the sum of the series 1 2 2 2 3 2 4 2 2n 2 Mathematics Stack Exchange Asked 10 years 7 months ago Modified 10 years 7 months ago Viewed 4k times 3 Find the sum of the series 12 22 32 42 2n 2 I tried rewriting it as 2n r 1 1n 1 r2 but it didn t help Recursion series Share Improve this question Follow asked Aug 10 2016 at 14 01 Kevin Patel 21 1 3 Add a comment 1 Answer Sorted by 0 Probably belongs to math section you can write the series as f 3k f 3k 1 1 f 3k 1
begingroup n 1 3 n 3 3n 2 3n 1 so it is clear that the n 2 terms can be added with some lower order terms attached by adding the differences of cubes giving a leading term in n 3 The factor 1 3 attached to the n 3 term is also obvious from this observation In mathematics 1 2 3 4 is an infinite series whose terms are the successive positive integers given alternating signs Using sigma summation notation the sum of the first m terms of the series can be expressed as
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1 5x 3 3 3x 4 4 2x 9 3 0 2 X 4 6x 9x x 4 2 3 256rzrss
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1 2 2 3 3 4 series - begin equation sum limits n 1 infty frac left 1 right n 1 n 1 frac 1 2 frac 1 3 frac 1 4 frac 1 5 frac 1 6 frac 1 7 frac 1 8 cdots ln 2 label eq eq1 end equation