English

Write the Sum of 20 Terms of the Series 1 + 1 2 ( 1 + 2 ) + 1 3 ( 1 + 2 + 3 ) + . . . . - Mathematics

Advertisements
Advertisements

Question

Write the sum of 20 terms of the series \[1 + \frac{1}{2}(1 + 2) + \frac{1}{3}(1 + 2 + 3) + . . . .\]

Solution

Let the nth term be \[a_n\]

Here, 

\[a_n = \frac{1}{n}\left( 1 + 2 + 3 + . . . + n \right) = \left( \frac{n + 1}{2} \right)\]

We know:

\[S_n = \sum^n_{k = 1} a_k\]

Thus, we have:

\[S_{20} = \sum^{20}_{k = 1} a_k\]

\[= \frac{1}{2}\left[ \sum^{20}_{k = 1} \left( k + 1 \right) \right]\]

\[ = \frac{1}{2}\left[ \sum^{20}_{k = 1} k + 20 \right]\]

\[ = \frac{1}{2}\left[ \frac{20\left( 21 \right)}{2} + 20 \right]\]

\[ = \frac{1}{2}\left[ 230 \right]\]

\[ = 115\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 21: Some special series - Exercise 21.3 [Page 19]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 21 Some special series
Exercise 21.3 | Q 6 | Page 19

Video TutorialsVIEW ALL [1]

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/(1xx2) + 1/(2xx3)+1/(3xx4)+ ...`


Find the sum to n terms of the series  52 + 62 + 72 + ... + 202


Find the sum to n terms of the series whose nth terms is given by n2 + 2n


Find the sum to n terms of the series whose nth terms is given by (2n – 1)2


1.2.4 + 2.3.7 +3.4.10 + ...


1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...


3 × 12 + 5 ×22 + 7 × 32 + ...


Find the sum of the series whose nth term is:

2n2 − 3n + 5


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 20th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 + ...


1 + 3 + 7 + 13 + 21 + ...


1 + 3 + 6 + 10 + 15 + ...


1 + 4 + 13 + 40 + 121 + ...


4 + 6 + 9 + 13 + 18 + ...


2 + 4 + 7 + 11 + 16 + ...


\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . .\]


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 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


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


The sum of 10 terms of the series \[\sqrt{2} + \sqrt{6} + \sqrt{18} +\] .... is

 

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


2 + 5 + 10 + 17 + 26 + ...

 

Find the natural number a for which ` sum_(k = 1)^n f(a + k)` = 16(2n – 1), where the function f satisfies f(x + y) = f(x) . f(y) for all natural numbers x, y and further f(1) = 2.


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 ______.


Let Sn(x) = `log_a  1/2 x + log_a  1/3 x + log_a  1/6 x + log_a  1/11 x  +  log_a  1/18 x + log_a  1/27x  + ` ...... up to n-terms, where a > 1. If S24(x) = 1093 and S12(2x) = 265, then value of a is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×