हिंदी

The Sum of the Series 2 3 + 8 9 + 26 27 + 80 81 + to N Terms is - Mathematics

Advertisements
Advertisements

प्रश्न

The sum of the series \[\frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{80}{81} +\] to n terms is

विकल्प

  • \[n - \frac{1}{2}( 3^{- n} - 1)\]

  • \[n - \frac{1}{2}(1 - 3^{- n} )\]

  • \[n + \frac{1}{2}( 3^n - 1)\]

  • \[n - \frac{1}{2}( 3^n - 1)\]

MCQ

उत्तर

\[n - \frac{1}{2}(1 - 3^{- n} )\]

Let

\[T_n\] be the nth term of the given series.
Thus, we have:

\[T_n = \frac{3^n - 1}{3^n} = 1 - \frac{1}{3^n}\]

Now,
Let

\[S_n\] be the sum of n terms of the given series.
Thus, we have:

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

\[ = \sum^n_{k = 1} \left[ 1 - \frac{1}{3^k} \right]\]

\[ = \sum^n_{k = 1} 1 - \sum^n_{k = 1} \frac{1}{3^k}\]

\[ = n - \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]

\[ = n - \frac{1}{3}\left[ \frac{1 - \left( \frac{1}{3} \right)^n}{1 - \frac{1}{3}} \right]\]

\[ = n - \frac{1}{2}\left[ 1 - \left( \frac{1}{3} \right)^n \right]\]

\[ = n - \frac{1}{2}\left[ 1 - 3^{- n} \right]\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 21: Some special series - Exercise 21.4 [पृष्ठ २०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 21 Some special series
Exercise 21.4 | Q 10 | पृष्ठ २०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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 whose nth term is given by n (n + 1) (n + 4).


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 + ...


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:

n (n + 1) (n + 4)


Find the sum of the series whose nth term is:

(2n − 1)2


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


3 + 7 + 14 + 24 + 37 + ...


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


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


The value of  \[\sum^n_{r = 1} \left\{ (2r - 1) a + \frac{1}{b^r} \right\}\] is equal to


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


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


The sum of the series

\[\frac{1}{\log_2 4} + \frac{1}{\log_4 4} + \frac{1}{\log_8 4} + . . . . + \frac{1}{\log_2^n 4}\] is


Sum of n terms of the series \[\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} +\] .......  is


The sum of the series 12 + 32 + 52 + ... to n terms is 


If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then write the value of k.


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 n terms


Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + ... to 10 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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×