## Ex 2.9 | Aster Classes

Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.9,

Question 1.
Find the sum of the following series
(i) 1 + 2 + 3 + …….. + 60
(ii) 3 + 6 + 9 + …….. +96
(iii) 51 + 52 + 53 + …….. + 92
(iv) 1 + 4 + 9 + 16 + …….. + 225
(v) 62 + 72 + 82 + …….. + 212
(vi) 103 + 113 + 123 + …….. + 203
(vii) 1 + 3 + 5 + …… + 71
Solution:
(i) 1 + 2 + 3 + …….. + 60 = 60×612
[Using n(n+1)2 formula]
= 1830.

(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)
= 3×32×332
= 1584

(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)

= 4278 – 1275
= 3003

(iv) 1 + 4 + 9 + 16 + …….. + 225 = 12 + 22 + 3+ 42 + ………… + 152
15×16×316
[using n(n+1)(2n+1)6] formula
= 1240

(v) 62 + 72 + 82 + …….. + 212 = 1 + 22 + 32 + 42 + ………… + 212 – (1 + 22 + ………… + 52)
= 21×22×436 – 5×6×116
= 3311 – 55
= 3256.

(vi) 103 = 113 + 123 + …….. + 203 = 13 + 23+ 33 + ………… + 203 – (13 + 23 + 33 + …………. + 93)

[Using (n(n+1)2)2 formula]
= 2102 – 452 = 44100 – 2025
= 42075

(vii) 1 + 3 + 5+ … + 71
Here a = 1; d = 3 – 1 = 2; l = 71

Question 2.
If 1 + 2 + 3 + …. + k = 325 , then find 13 + 23+ 33 + …………. + k3
1 + 2 + 3 + …. + k = 325
k(k+1)2 = 325 ……(1)

= 3252 (From 1)
= 105625.

Question 3.
If 13 + 23 + 33 + ………… + K3 = 44100 then find 1 + 2 + 3 + ……. + k
13 + 23 + 33 + ………….. + k3 = 44100

k(k+1)2 = 44100−−−−−√ = 210
1 + 2 + 3 + …… + k = k(k+1)2
= 210

Question 4.
How many terms of the series 13 + 23 + 33+ …………… should be taken to get the sum 14400?
13 + 23 + 33 + ……. + n3 = 14400

n(n+1)2 = 14400−−−−−√
n(n+1)2 = 120 ⇒ n2 + n = 240

n2 + n – 240 = 0
(n + 16) (n – 15) = 0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15

Question 5.
The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
12 + 22 + 32 + …. + n2 = 285

Question 6.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated with these colour papers?
Area of 15 square colour papers
= 102 + 112 + 122 + …. + 242
= (12 + 22 + 32 + …. + 242) – (12 + 22 + 92)
= 24×25×496−9×10×196
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm2

Question 7.
Find the sum of the series (23 – 1)+(43 – 33) + (63 – 153) + …….. to
(i) n terms
(ii) 8 terms
Sum of the series = (23 – 1) + (43 – 33) + (63 – 153) + …. n terms
= 23 + 43 + 63 + …. n terms – (13 + 33 + 53+ …. n terms) …….(1)
23 + 43 + 63 + …. n = ∑(23 + 43 + 63 + ….(2n)3]
∑ 23 (13 + 23 + 33 + …. n3)
= 8 (n(n+1)2)2
= 2[n (n + 1)]2
13 + 33 + 53 + ……….(2n – 1)3 [sum of first 2n cubes – sum of first n even cubes]

Substituting (2) and (3) in (1)
Sum of the series = 2n2 (n + 1)2 – n2 (2n + 1)2 + 2n2(n + 1)2
= 4n2 (n + 1)2 – n2 (2n + 1)2
= n2 [(4(n + 1)2 – (2n + 1)2]
= n2 [4n2 + 4 + 8n – 4n2 – 1 – 4n]
= n2 [4n + 3]
= 4n3 + 3n2

(ii) when n = 8 = 4(8)3 + 3(8)2
= 4(512) + 3(64)
= 2240.

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