RD Chapter 21 Some Special Series Ex 21.1 Solutions
Question - 1 : - Find the sum of the following series to n terms: 13 + 33 + 53 +73 + ……..
Answer - 1 : -
Let Tn bethe nth term of the given series.
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
Now, let Sn bethe sum of n terms of the given series.
We have:
Upon simplification weget,
= 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]
∴ The sum of the seriesis n2 [2n2 – 1]
23 + 43 + 63 + 83 +………
Answer - 2 : -
Let Tn bethe nth term of the given series.
We have:
Tn =(2n)3
= 8n3
Now, let Sn bethe sum of n terms of the given series.
We have:
∴ The sum of the seriesis 2{n (n + 1)}2
Question - 3 : - 1.2.5 + 2.3.6 + 3.4.7 + ……..
Answer - 3 : -
Let Tn bethe nth term of the given series.
We have:
Tn = n(n + 1) (n + 4)
= n (n2 +5n + 4)
= n3 +5n2 + 4n
Now, let Sn bethe sum of n terms of the given series.
We have:
Question - 4 : - 1.2.4 + 2.3.7 + 3.4.10 + … to n terms.
Answer - 4 : -
Let Tn bethe nth term of the given series.
We have:
Tn = n(n + 1) (3n + 1)
= n (3n2 +4n + 1)
= 3n3 +4n2 + n
Now, let Sn bethe sum of n terms of the given series.
We have:
∴ The sum of the seriesis
Question - 5 : - 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + … to n terms
Answer - 5 : -
Let Tn bethe nth term of the given series.
We have:
Tn =n(n+1)/2
= (n2 +n)/2
Now, let Sn bethe sum of n terms of the given series.
We have:
∴ The sum of the seriesis [n(n+1)(n+2)]/6