Class 11: Special Series – Exercise 21.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Notes:

$\displaystyle S_n = \sum \limits_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$
$\displaystyle S_n = \sum \limits_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$
$\displaystyle S_n = \sum \limits_{k=1}^{n} k^2 = \frac{n ( n+1)(2n+1)}{6}$
$\displaystyle S_n = \sum \limits_{k=1}^{n} k = \frac{n(n+1)}{2}$
$\displaystyle S_n = \sum \limits_{k=1}^{n} 1 = n$

Find the sum of the following series to $\displaystyle n$ terms:

Question 1: $\displaystyle 1^3 + 3^3 + 5^3 + 7^3 + \ldots$

Answer:

Given series: $\displaystyle 1^3 + 3^3 + 5^3 + 7^3 + \ldots$

$\displaystyle = ( 2 \times 1 -1)^3 + ( 2 \times 2 -1)^3 + ( 2 \times 3 -1)^3 + ( 2 \times 4 -1)^3 + \ldots$

$\displaystyle \therefore T_n = (2n-1)^3$

$\displaystyle \therefore S_n = \sum \limits_{k=1}^{n} (2n-1)^3$

$\displaystyle = \sum \limits_{k=1}^{n} (8k^3 - 1 - 12k^2 + 6k)$

$\displaystyle = 8 \sum \limits_{k=1}^{n} k^3 - 12 \sum \limits_{k=1}^{n}k^2 + 6 \sum \limits_{k=1}^{n} - \sum \limits_{k=1}^{n} 1$

$\displaystyle = \frac{8n^2(n+1)^2}{4} - \frac{12n ( n+1)(2n+1)}{6} + \frac{6n(n+1)}{2} - n$

$\displaystyle = 2n^2( n+1)^2 - 2n (n+1)(2n+1) + 3n(n+1)-n$

$\displaystyle = n(n+1)[ 2n(n+1) - 2 ( 2n+1) + 3 ] - n$

$\displaystyle = n(n+1)[ 2n^2 + 2n - 4n - 2 - 3] - n$

$\displaystyle = n ( n+1) [ 2n^2 - 2n +1] - n$

$\displaystyle = n [ (n+1)(2n^2 - 2n +1) - 1]$

$\displaystyle = n [ 2n^3 - 2n^2 + n + 2n^2 - 2n + 1 - 1]$

$\displaystyle = n ( 2n^3 - n)$

$\displaystyle = n^2 ( 2n^3 - 1)$

$\displaystyle \\$

Question 2: $\displaystyle 2^3 + 4^3 + 6^3 + 8^3 + \ldots$

Answer:

Given series: $\displaystyle 2^3 + 4^3 + 6^3 + 8^3 + \ldots$

$\displaystyle = (2 \times 1)^2 + (2 \times 2)^2 + (2 \times 3)^2 + (2 \times 4)^2 +\ldots$

$\displaystyle \therefore T_n = (2n)^2$

$\displaystyle \therefore S_n = \sum \limits_{k=1}^{n} (2k)^2$

$\displaystyle = 4 \sum \limits_{k=1}^{n} k^2$

$\displaystyle = \frac{4n(n+1)(2n+1)}{6}$

$\displaystyle = \frac{2}{3} n (n+1)(2n+1)$

$\displaystyle \\$

Question 3: $\displaystyle 1 \cdot 2 \cdot 5 + 2 \cdot 3 \cdot 6 + 3 \cdot 4 \cdot 7 + \ldots$

Answer:

Given series: $\displaystyle 1 \cdot 2 \cdot 5 + 2 \cdot 3 \cdot 6 + 3 \cdot 4 \cdot 7 + \ldots$

$\displaystyle = (1)(1+1)(1+4) + (2)(2+1)(2+4) + (3)(3+1)(3+4) + (4)(4+1)(4+4) + \ldots$

$\displaystyle \therefore T_n = n(n+1)(n+4) = n(n^2+5n+4) = n^3 + 5n^2 + 4n$

$\displaystyle \therefore S_n = \sum \limits_{k=1}^{n} (k^3 + 5k^2 + 4k)$

$\displaystyle = \sum \limits_{k=1}^{n} k^3 + 5 \sum \limits_{k=1}^{n} k^2+ 4 \sum \limits_{k=1}^{n} k$

$\displaystyle = \frac{n^2(n+1)^2}{4} + \frac{5n ( n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{n(n+1)}{2} + \frac{5(2n+1)}{3} + 4 \Big]$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{n^2+n}{2} + \frac{10n+5}{3} + 4 \Big]$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{3n^2+3n+20n+10+24}{6} \Big]$

$\displaystyle = \frac{n(n+1)}{2} (3n^2+23n+34)$

$\displaystyle \\$

Question 4: $\displaystyle 1 \cdot 2 \cdot 4 + 2 \cdot 3 \cdot 7 + 3 \cdot 4 \cdot 10 + \ldots$

Answer:

Given series: $\displaystyle 1 \cdot 2 \cdot 4 + 2 \cdot 3 \cdot 7 + 3 \cdot 4 \cdot 10 + \ldots$

$\displaystyle = 1(1+1)(3 \times 1 + 1) + 2(2+1)(3 \times 2 + 1) + 3(3+1)(3 \times 3 + 1) + 4(4+1)(3 \times 4 + 1) + \ldots$

$\displaystyle \therefore T_n = n(n+1)(3n+1) = n ( 3n^2 + 3n + n + 1) = 3n^3 + 4n^2 + n$

$\displaystyle \therefore S_n = \sum \limits_{k=1}^{n} (3k^3 + 4k^2 + k)$

$\displaystyle = 3 \sum \limits_{k=1}^{n} k^3 + 4 \sum \limits_{k=1}^{n} k^2 + \sum \limits_{k=1}^{n} k$

$\displaystyle = \frac{3n^2(n+1)^2}{4} + \frac{4n ( n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$\displaystyle = \frac{3n^2(n+1)^2}{4} + \frac{2n ( n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{3n(n+1)}{2} + \frac{4(2n+1)}{3} + 1 \Big]$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{3n^2+3n}{2} + \frac{8n+4}{3} + 1 \Big]$

$\displaystyle = \frac{n(n+1)}{2} \Big [ \frac{9n^2+9n+16n+8+6}{6} \Big ]$

$\displaystyle = \frac{n(n+1)}{2} (9n^2+25n+14)$

$\displaystyle \\$

Question 5: $\displaystyle 1 + ( 1+2) + ( 1 + 2 + 3) + ( 1 + 2 + 3 + 4) + \ldots$

Answer:

Given series: $\displaystyle 1 + ( 1+2) + ( 1 + 2 + 3) + ( 1 + 2 + 3 + 4) + \ldots$

$\displaystyle \therefore T_n = 1 + 2 + 3 + 4 + \ldots + n = \frac{n(n+1)}{2} = \frac{n^2 + n}{2}$

$\displaystyle \therefore S_n = \frac{1}{2} \sum \limits_{k=1}^{n} k^2 + \frac{1}{2} \sum \limits_{k=1}^{n} k$

$\displaystyle = \frac{1}{2} \Big[ \frac{n ( n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \Big]$

$\displaystyle = \frac{n(n+1)}{4} \Big[ \frac{2n+1}{3} + 1 \Big]$

$\displaystyle = \frac{n(n+1)}{4} \Big[ \frac{2n+4}{3} \Big]$

$\displaystyle = \frac{1}{12} n ( n+1)(2n+4)$

$\displaystyle \\$

Question 6: $\displaystyle 1 \times 2 +2 \times 3 +3 \times 4 +5 \times 5 + \ldots$

Answer:

Given series: $\displaystyle 1 \times 2 +2 \times 3 +3 \times 4 +5 \times 5 + \ldots$

$\displaystyle \therefore T_n = n(n+1) = n^2 + n$

$\displaystyle \therefore S_n = \sum \limits_{k=1}^{n} (k^2+k)$

$\displaystyle = \sum \limits_{k=1}^{n} k^2 + \sum \limits_{k=1}^{n} k$

$\displaystyle = \frac{n ( n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$\displaystyle = \frac{n(n+1)}{2} \Big [ \frac{2n+1}{3} + 1 \Big]$

$\displaystyle = \frac{n(n+1)}{2} \frac{(2n+4)}{3}$

$\displaystyle = \frac{1}{3} n ( n+1)(n+2)$

$\displaystyle \\$

Question 7: $\displaystyle 3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + \ldots$

Answer:

Given series: $\displaystyle 3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + \ldots$

$\displaystyle \therefore T_n = (2n+1)(n^2) = 2n^3 + n^2$

$\displaystyle \therefore S_n = \sum \limits_{k=1}^{n} (2k^3+k^2)$

$\displaystyle = 2 \sum \limits_{k=1}^{n} k^3 + \sum \limits_{k=1}^{n} k^2$

$\displaystyle = \frac{2n^2(n+1)^2}{4} + \frac{n ( n+1)(2n+1)}{6}$

$\displaystyle = \frac{n^2(n+1)^2}{2} + \frac{n ( n+1)(2n+1)}{6}$

$\displaystyle = \frac{n(n+1)}{2} \Big[ n(n+1) + \frac{2n+1}{3} \Big]$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{3n^2+3n+2n+1}{3} \Big]$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{3n^2+5n+1}{3} \Big]$

$\displaystyle = \frac{1}{6} n(n+1)(3n^2+5n+1)$

$\displaystyle \\$

Question 8: Find the sum of the series whose $latex \displaystyle n^{th} term is:

i) $\displaystyle 2n^3+3n^2-1$ ii) $\displaystyle n^3-3^n$ iii) $\displaystyle n ( n+1) (n+4)$ iv) $\displaystyle (2n-1)^2$

Answer:

i) Given $\displaystyle T_n = 2n^3+3n^2-1$

$\displaystyle S_n = \sum \limits_{k=1}^{n} ( 2k^3+3k^2-1)$

$\displaystyle = 2 \sum \limits_{k=1}^{n} k^3 + 3 \sum \limits_{k=1}^{n} k^2 - \sum \limits_{k=1}^{n} 1$

$\displaystyle = \frac{2n^2(n+1)^2}{4} + \frac{3n ( n+1)(2n+1)}{6} - n$

$\displaystyle = \frac{n^2(n+1)^2}{2} + \frac{n ( n+1)(2n+1)}{2} - n$

$\displaystyle = \frac{n(n+1)}{2} [ n(n+1) + (2n+1)] - n$

$\displaystyle = \frac{n(n+1)}{2} [ n^2 + 3n + 1] - n$

$\displaystyle = \frac{1}{2} [ (n^2+n)(n^2 + 3n+1) - 2n]$

$\displaystyle = \frac{1}{2} [ n^4 + n^3 + 3n^3 + 3n^2 + n^2 + n - 2n ]$

$\displaystyle = \frac{1}{2} [ n^4 + 4n^3 + 4n^2 - n ]$

$\displaystyle = \frac{n}{2} [ n^3 + 4n^2 + 4n - 1]$

ii) Given $\displaystyle T_n = n^3-3^n$

$\displaystyle S_n = \sum \limits_{k=1}^{n} k^3 - 3^k$

$\displaystyle = \sum \limits_{k=1}^{n} k^3 - \sum \limits_{k=1}^{n} 3^k$

$\displaystyle = \frac{n^2(n+1)^2}{4} - (3^1 + 3^2 + 3^3 + \ldots + 3^n)$

$\displaystyle = \frac{n^2(n+1)^2}{4} - \frac{3(3^n -1)}{3-1}$

$\displaystyle = \frac{n^2(n+1)^2}{4} - \frac{3}{2} (3^n -1)$

iii) Given $\displaystyle T_n = n ( n+1) (n+4) = n^3 + 5n^2 + 4n$

$\displaystyle S_n = \sum \limits_{k=1}^{n} (k^3 + 5k^2 + 4k)$

$\displaystyle = \sum \limits_{k=1}^{n} k^3 + 5 \sum \limits_{k=1}^{n} k^2 + 4 \sum \limits_{k=1}^{n} k$

$\displaystyle = \frac{n^2(n+1)^2}{4} + \frac{5n ( n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{n(n+1)}{2} + \frac{5(2n+1)}{3} + 4 \Big]$

$\displaystyle = \frac{n(n+1)}{2} \Big [ \frac{n^2 + n}{2} + \frac{10n+5}{3} + 4 \Big]$

$\displaystyle = \frac{n(n+1)}{12} \Big [ 3n^2 + 3n + 20 n + 10 + 24 \Big]$

$\displaystyle = \frac{n(n+1)}{12} \Big [ 3n^2 + 23n + 34 \Big]$

iv) Given $\displaystyle T_n = (2n-1)^2 = 4n^2 + 1 - 4n$

$\displaystyle S_n = \sum \limits_{k=1}^{n} ( 4k^2 + 1 - 4k)$

$\displaystyle = 4 \sum \limits_{k=1}^{n} k^2 + \sum \limits_{k=1}^{n} 1 - 4 \sum \limits_{k=1}^{n} k$

$\displaystyle = \frac{4n ( n+1)(2n+1)}{6} + n - \frac{4n(n+1)}{2}$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{4(2n+1)}{3} - 4 \Big] + n$

$\displaystyle = \frac{n(n+1)}{2} \Big[ \frac{8n+4 - 12}{3} \Big] + n$

$\displaystyle = \frac{n(n+1)}{2} \Big [ \frac{8n -8}{3} \Big] + n$

$\displaystyle = \frac{4n(n+1)(n-1)}{3} + n$

$\displaystyle = \frac{4n ( n^2 - 1) + 3n}{3}$

$\displaystyle = \frac{4n^3 - 4n+ 3n}{3}$

$\displaystyle = \frac{4n^3 - n}{3}$

$\displaystyle = \frac{1}{3} n ( 2n - 1)(2n +1)$

$\displaystyle \\$

Question 9: Find the $\displaystyle 20^{th}$ term and the sum of $\displaystyle 20$ terms of the series :

$\displaystyle 2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots$

Answer:

Given series: $\displaystyle 2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots$

$\displaystyle = ( 2 \times 1)( 2 \times 1 + 2) + ( 2 \times 2)( 2 \times 2 + 2) + ( 2 \times 3)( 2 \times 3 + 2) + \ldots$

$\displaystyle \therefore T_n = 2n ( 2n+2) = 4n^2 + 4n$

$\displaystyle S_n = \sum \limits_{k=1}^{n} (4k^2 + 4k)$

$\displaystyle = 4 \sum \limits_{k=1}^{n} k^2 + 4 \sum \limits_{k=1}^{n} k$

$\displaystyle = \frac{4n ( n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$

$\displaystyle \therefore S_{20} = \frac{4(20) ( 20+1)(2 \times 20+1)}{6} + \frac{4(20)(20+1)}{2} = 11480 + 840 = 12320$

Also $\displaystyle T_{20} = 4 (20)^2 + 4(20) = 4 \times 400 + 80 = 1680$

$\displaystyle \\$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 11: Special Series – Exercise 21.1

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