CBSE Class 11 Mathematics Sequences And Series Worksheet Set 06

CBSE Class 11 Mathematics Worksheet - Sequences and Series (5). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Find the sum to \( n \) terms \( 1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 \dots\dots\dots\dots n \) terms.
Answer:
The general term of this series is given by:
\( a_n = (n)(n + 1)(n + 2) \)
\( a_n = n(n^2 + 3n + 2) \)
\( a_n = n^3 + 3n^2 + 2n \)

Now, \( S_n = \sum a_n \)
\( = \sum (n^3 + 3n^2 + 2n) \)
\( = \sum n^3 + 3 \sum n^2 + 2 \sum n \)
\( = \frac{n^2(n+1)^2}{4} + \frac{3n(n+1)(2n+1)}{6} + \frac{2n(n+1)}{2} \)
\( = n(n + 1) \left[ \frac{n(n+1)}{4} + \frac{2n+1}{2} + 1 \right] \)
\( = n(n + 1) \left[ \frac{n^2+n+4n+2+4}{4} \right] \)
\( = \frac{n(n+1)(n^2+5n+6)}{4} \)
\( S_n = \frac{n(n+1)(n+2)(n+3)}{4} \) ans.

Question. Find the sum to \( n \) terms \( 3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + \dots\dots\dots\dots n \) terms.
Answer:
General term of this series is:
\( a_n = (2n + 1)(n^2) \)
\( a_n = (2n^3 + n^2) \)

Now, \( S_n = \sum a_n \)
\( = \sum(2n^3 + n^2) \)
\( = 2 \sum n^3 + \sum n^2 \)
\( = 2 \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6} \)
\( = n(n + 1) \left[ \frac{n(n+1)}{2} + \frac{2n+1}{6} \right] \)
\( = n(n + 1) \left[ \frac{3n(n+1) + (2n+1)}{6} \right] \)
\( = n(n + 1) \left[ \frac{3n^2+3n+2n+1}{6} \right] \)
\( = \frac{n(n+1)(3n^2+5n+1)}{6} \)
\( S_n = \frac{n(n+1)(3n^2+5n+1)}{6} \) ans.

Question. Find the sum \( 5^2 + 6^2 + 7^2 \dots\dots\dots 20^2 \).
Answer:
Let \( S = 5^2 + 6^2 + 7^2 \dots\dots\dots 20^2 \).
\( S = (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 \dots\dots\dots 20^2) - (1^2 + 2^2 + 3^2 + 4^2) \)
\( S = \left[ \frac{n(n+1)(2n+1)}{6} \right] - \left[ \frac{n(n+1)(2n+1)}{6} \right] \)

Here, put \( n = 20 \) for the first term, and \( n = 4 \) for the second term:
\( S = \left[ \frac{20(20+1)(40+1)}{6} \right] - \left[ \frac{4(4+1)(8+1)}{6} \right] \)
\( S = \frac{20(21)(41)}{6} - \frac{4(5)(9)}{6} \)
\( = 2870 - 30 = 2840 \) ans.

Question. Find the sum to \( n \) terms \( \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots\dots\dots\dots n \) terms.
Answer:
General term of the series is:
\( a_n = \frac{1}{n(n+1)} \)

Let \( S_n = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots\dots + \frac{1}{n(n+1)} \)
\( S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots\dots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \)
\( S_n = 1 - \frac{1}{n+1} \)
\( S_n = \frac{n+1-1}{n+1} \)
\( \therefore S_n = \frac{n}{n+1} \) ans.