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Class 11 RD Sharma Solutions – Chapter 21 Some Special Series- Exercise 21.1

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Find the sum of the following series to n terms.(Question 1-5)

Question 1. 13 + 33 + 53 + 73 + ……

Solution:

Let us assume Tn to be the nth term of the given series.

Now, we have:

Tn = [1 + (n – 1)2]3

= (2n – 1)3

= (2n)3 – 3 (2n)2. 1 + 3.12.2n – 13  [Since, (a – b)3 = a3 – 3a2b + 3ab2 – b]

= 8n3 – 12n2 + 6n – 1

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^n T_k \\ =  \sum_{k=1}^n [2k-1]^3 \\ =  \sum_{k=1}^n [8k^3 - 1 - 6k(2k-1)] \\ =  \sum_{k=1}^n [8k^3 - 1 - 12k^2 + 6k] \\ =  8 \sum_{k=1}^n k^3 -  \sum_{k=1}^n 1 - 12  \sum_{k=1}^n k^2 + 6 \sum_{k=1}^n k \\ = \frac{8n^2(n+1)^2}{4} - n - \frac{12n(n+1)(2n+1)}{6} + \frac{6n(n+1)}{2}

Simplifying the equation we get

= 2n2 (n + 1)2 – n – 2n (n + 1) (2n + 1) + 3n (n + 1)

= n (n + 1) [2n (n + 1) – 2 (2n + 1) + 3] – n

= n (n + 1) [2n2 – 2n + 1] – n

= n [2n3 – 2n2 + n + 2n2 – 2n + 1 – 1]

= n [2n3 – n]

= n2 [2n2 – 1]

Therefore, 

The sum of the series is n2 [2n2 – 1].

Question 2. 23 + 43 + 63 + 83 + ………

Solution:

Let us assume Tn to be the nth term of the given series.

Now, we have:

Tn = (2n)3

= 8n3

Also, let us assume Sn to be the sum of n terms of the given series.

S_n =  \sum_{k=1}^n 8k^3 \\ = 8  \sum_{k=1}^n k^3 \\ = 8[\frac{n(n+1)}{2}]^2 \\ = 8 * \frac{n^2(n+1)^2}{4} \\ = 2n^2(n+1)^2 \\ = 2{n(n+1)}^2

Therefore, 

The sum of the given series is 2{n (n + 1)}2

Question 3. 1.2.5 + 2.3.6 + 3.4.7 + ……

Solution:

Let us assume Tn to be the nth term of the given series.

Now, we have:

Tn = n (n + 1) (n + 4)

= n (n2 + 5n + 4)

= n3 + 5n2 + 4n

Also, let us assume Sn to be the sum of n terms of the given series.

S_n =  \sum_{k=1}^n T_k \\  \sum_{k=1}^n k^3 +  \sum_{k=1}^n 5k^2 +  \sum_{k=1}^n4k \\ =  \sum_{k=1}^nk^3 + 5  \sum_{k=1}^nk^2 + 4  \sum_{k=1}^n \\ = \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2} \\ = \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + 2n(n+1) \\ = \frac{n(n+1)}{2} (\frac{n(n+1)}{2} + \frac{5(2n+1)}{3} + 4 ) \\ = \frac{n(n+1)}{2} (\frac{n^2 + n}{2} + \frac{10n+5}{3} + 4 ) \\ = \frac{n(n+1)}{2} (\frac{3n^2 + 3n + 20n + 10 + 24}{6}) \\ = \frac{n}{12} (n+1)(3n^2+23n+34) \\ = \frac{n(n+1)}{2}(\frac{3n^2+3n+20n+10+24}{6}) \\ = \frac{n}{12} (n+1) (3n^2 + 23n + 34)

Therefore, 

The sum of the given series is \frac{n}{12}(n+1)(3n^2 + 23n + 34)

Question 4. 1.2.4 + 2.3.7 + 3.4.10 + ….

Solution:

Let us assume Tn to be the nth term of the given series.

Now, we have:

Tn = n (n + 1) (3n + 1)

= n (3n2 + 4n + 1)

= 3n3 + 4n2 + n

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^n T_k \\ = \sum_{k=1}^n 3k^3 + \sum_{k=1}^n 4k^2 + \sum_{k=1}^n k \\ = 3\sum_{k=1}^n k^3 + 4\sum_{k=1}^n k^2 + \sum_{k=1}^n k \\ = \frac{3n^2(n+1)^2}{4} + \frac{4n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ = \frac{3n^2(n+1)^2}{4} + \frac{2n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \\ = \frac{n(n+1)}{2} (\frac{3n(n+1)}{2} + \frac{4(2n+1)}{3} + 1 ) \\ = \frac{n(n+1)}{2} (\frac{3n^2+3n}{2} + \frac{8n+4}{3} + 1) \\ = \frac{n(n+1)}{2} (\frac{9n^2+9n+16n+8+6}{6}) \\ = \frac{n}{12}(n+1)(9n^2+25n+14)

Therefore, 

The sum of the given series is \frac{n}{12}(n+1)(9n^2 + 25n + 14)

Question 5. 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + …

Solution:

Let us assume Tn to be the nth term of the given series.

Now, we have:

Tn = n(n + 1)/2

= (n2 + n)/2

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^n T_k \\ = \sum_{k=1}^n (\frac{k^2+k}{2}) \\ = 1/2 \sum_{k=1}^n (k^2+k) \\ = 1/2 [\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ] \\ = \frac{n(n+1)}{4}(\frac{2n+1}{3} + 1) \\ = \frac{n(n+1)}{4} (\frac{2n+4}{3}) \\ = \frac{n(n+1)(2n+4)}{12} \\ = \frac{n(n+1)(n+2)}{6}

Therefore, 
The sum of the series is [n(n + 1)(n + 2)]/6

Question 6. Find the sum of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … upto n terms.

Solution:

Let us assume Tn to be the nth term of the given series.
Tn = (nth term of 1, 2, 3..) x (nth term of 2, 3, 4…)

= [1 + (n + 1) x 1].[2 + (n + 1) x 1]

= [1 + n – 1].[2 + n – 1]

= n(n + 1)

= n2 + n 

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^n T_n \\ = \sum_{k=1}^n (n^2 + n) \\ = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ = \frac{n(n+1)(2n+1)+3n(n+1)}{6} \\ = \frac{n(n+1)[2n+1+3]}{6} \\ = \frac{n(n+1)[2n+4]}{6} \\ = \frac{n(n+1) x 2 (n+2)}{6} \\ = \frac{n}{6} (n+1)(n+2)

Question 7. Find the sum of the series 3 × 12 + 5 × 22 + 7 × 32 + … upto n terms.

Solution:

Let us assume Tn to be the nth term of the given series.

Tn = (nth term of 3, 5, 7..) x (nth term of 12, 22, 32…)

= [3 + (n – 1) x 2].[n2]

= [2n + 1]. [n2]

= 2n3 + n2

Tn = 2n3 + n2

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^n T_n \\ = \sum_{k=1}^n (2n^3 + n^2) \\ = \sum_{k=1}^n 2n^3 + \sum_{k=1}^n n^2 \\ = 2 \sum_{k=1}^n n^3 + \sum_{k=1}^nn^2 \\ = 2[\frac{n(n+1)}{2}]^2 + [\frac{n(n+1)(2n+1)}{6}] \\ = \frac{2}{4}[n(n+1)]^2 + \frac{n(n+1)(2n+1)}{6} \\ = [\frac{n(n+1)}{2}]^2 + \frac{n(n+1)(2n+1)}{6} \\ = \frac{n(n+1)}{6}[3n(n+1)+(2n+1)] \\ = \frac{n(n+1)}{6}[3n^2+3n+2n+1] \\ \frac{n}{6}(n+1)(3n^2+5n+1)

Therefore, 

The sum of the series = \frac{n}{6}(n+1)(3n^2+5n+1)

Question 8 (i). Find the sum of the series 2n3 + 3n2 – 1 to n terms.

Solution:

Tn = 2n3 + 3n2 – 1 

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^{n}T_k \\ = 2 \sum_{k=1}^{n} 2k^3+ \sum_{k=1}^{n}3k^2 - \sum_{k=1}^{n} 1 \\ = 2[\frac{n(n+1)}{2}]^2 + 3[\frac{n(n+1)(2n+1)}{6}] -n \\ = \frac{2}{4}[n(n+1)]^2 + \frac{n(n+1)(2n+1) - n }{2} \\ = \frac{[n(n+1]^2+n(n+1)(2n+1)-n}{2} \\ = \frac{n}{2}[n(n+1)^2 + (n+1)(2n+1) - 2] \\ = \frac{n}{2} [n^3+n+2n^2+2n^2+3n-1] \\ = \frac{n}{2} [n^3+4n^2+4n-1]

Sum of n terms = \frac{n}{2} [n^3+4n^2+4n-1]

Question 8 (ii). Find the sum of the series n3 – 3n to n terms.

Solution:

Tn = n3 – 3n

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^{n}T_k \\ =  \sum_{k=1}^{n}(k^3-3^k) \\ = \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} 3^k \\ = [\frac{n(n+1)}{2}]^2-(3^1+3^2+...+3^n) \\ = \frac{n^2(n+1)^2}{4}-3\frac{3^n-1}{3-1} \\ = \frac{n^2(n+1)^2}{4}-\frac{3}{2}(3^n-1)

Question 8 (iii). Find the sum of the series n(n + 1)(n + 4) to n terms.

Solution:

Tn = n(n + 1)(n + 4)

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^{n}T_k \\ = \sum_{k=1}^{n}(k^3+5k^2+4k) \\ = \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} 5k^2 + \sum_{k=1}^{n} 4k \\ = \sum_{k=1}^{n} k^3 + 5 \sum_{k=1}^{n} k^2 + 4 \sum_{k=1}^{n}k \\ = [\frac{n(n+1)}{2}]^2 + 5[\frac{n(n+10(2n+1)}{6}]+\frac{4n(n+1)}{2} \\ = \frac{1}{4}[n(n+1)]^2 + \frac{5n(n+1)(2n+1}{6}+2n(n+1) \\ = \frac{3[n(n+1)]^2+10n(n+1)(2n+1)+24n(n+1)}{12} \\ = \frac{n(n+1)}{12}[3n(n+1)+10(2n+1)+24] \\ = \frac{n(n+1)}{12}[3n^2+23n+34]

Question 8 (iv). Find the sum of the series (2n – 1)2 to n terms.

Solution:

Tn = (2n – 1)

(2n)^2 + 1 -2n \\ = 4n^2 + 1 - 4n \\ = 4n^2 - 4n + 1

Also, let us assume Sn to be the sum of n terms of the given series.

S_n = \sum_{k=1}^{n}T_k \\ = \sum_{k=1}^{n}4k +\sum_{k=1}^{n}1 \\ = \sum_{k=1}^{n}k^2 -4\sum_{k=1}^{n}k + \sum_{k=1}^{n}1 \\ = 4[\frac{n(n+1)(2n+1)}{6}]-4\frac{n(n+1)}{2}+n \\ = 2/3n(n+1)(2n+1)-2n(n+1)+n \\ = \frac{2}{3}n(n+1)(2n+1)-2n^2-2n+n \\ = \frac{2n}{3}(n+1)(2n+1) - n(2n+1) \\ = \frac{n}{3} (2n+1)[2(n+1)-3] \\ = \frac{n}{3}(2n+1)(2n+2-3) \\ = \frac{n}{3}(2n+1)(2n-1)

Question 9. Find the 20th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 +…..  

Solution:

Let us assume Tn to be the nth term of the given series.

T_n = 2n(2n+2)

The 20th term of the series is :

T_{20} = 2 * 20(2*20+2) = 1680

The infinite series is equivalent to, 

2 × 4 + 4 × 6 + 6 × 8 + …. = \sum_{n=1} 2n(2n+2)

Sum of the series until 20th term is equivalent to

\sum_{n=1}^{20} 2n (2n+2) = \sum_{n=1}^{20}4n^2 + \sum_{n=1}^{20}4n \\ = 4 \sum_{n=1}^{20} n^2 + 4\sum_{n=1}^{20} n \\ = 4\frac{20(20+1)(2.20+1)}{6}+4. \frac{20(20+1)}{2} \\ = 12320


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Last Updated : 18 Mar, 2021
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