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Sequences and Series — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSSequences and Series5 Marks
Question
(33 - 23) + (53 - 43) + (73 - 63) + .... to:
  1. n terms.
  2. 10 terms.
✓

Answer

Given series:
= (33 - 23) + (53 - 43) + (73 - 63) + ....
= (33 + 53 + 73 + ....) - (23 + 43 + 63 + ....)
= [33 + 53 + 73 + .... (2n + 1)3] - [23 + 43 + 63 + .... (2n)3]
$\therefore$ Tn = (2n + 1)3 - (2n)3
= (2n + 1 - 2n) [(2n + 1)2 + (2n + 1)(2n) + (2n)2] $[\because$ a3 - b3 = (a - b)(a2 + ab + b2)$]$
= 1 - [4n2 + 1 + 4n + 4n2 + 2n + 4n2]
=12n2 + 6n + 1
  1. $\text{S}_\text{n}=\sum\text{T}_\text{n}=12\sum\text{n}^2+6\sum\text{n}+\text{n}$

$=12.\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{6\text{n}(\text{n}+1)}{2}+\text{n}$

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

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

= n[2(2n2 + 3n + 1) + 3n + 3 + 1]

= n[4n2 + 6n + 2 + 3n + 4]

= n[4n2 + 9n + 6]

= 4n3 + 9n2 + 6n

  1. S10 = 4(10)3 + 9(10)2 + 6(10) = 4 × 1000 + 900 + 60

= 4000 + 960 = 4960

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