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Sequences and Series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSequences and Series5 Marks
Question
$(3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + ....$ to:
$(1) \ n$ terms.
$(2) \ 10$ terms.
✓

Answer

Given series:
$= (3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + ....$
$= (3^3 + 5^3 + 7^3 + ....) - (2^3+ 4^3 + 6^3+ ....)$
$= [3^3 + 5^3 + 7^3 + .... (2n + 1)^3] - [2^3 + 4^3 + 6^3 + .... (2n)^3]$
$\therefore T_n = (2n + 1)^3 - (2n)^3$
$= (2n + 1 - 2n) [(2n + 1)^2+ (2n + 1)(2n) + (2n)^2] [\because a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$= 1 - [4n^2 + 1 + 4n + 4n^2 + 2n + 4n^2]$
$=12n^2 + 6n + 1$
$\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}$
$= 2n(n + 1)(2n + 1) + 3n(n + 1) + n$
$= n[2(n + 1)(2n + 1) + 3n(n + 1) + 1]$
$= n[2(2n^2 + 3n + 1) + 3n + 3 + 1]$
$= n[4n^2 + 6n + 2 + 3n + 4]$
$= n[4n^2 + 9n + 6]$
$= 4n^3+ 9n^2 + 6n$
$S_{10} = 4(10)^3 + 9(10)^2 + 6(10) = 4 \times 1000 + 900 + 60= 4000 + 960 = 4960$

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