Find the sum of series 1 . 2 + 2 . 22 + 3 . 22 +…+ 100 . 2100.
Solution:
Let S = 1.2 + 2.22 + 3.23 +…+ 100.2100 …… (1)
⇒ 2S = 1.22 + 2.23 +…+ 99.2100 + 100.2101 …… (2)
⇒ –S = 1.2 + 1.22 + 1.23 +…+ 1.2100 – 100.2101
⇒ –S = 1.2 (2100–1/2–1) – 100.2101
⇒ S = –2101 + 2 + 100.2101 = 199.2101 + 2.
Illustration:
Let r = 1/2, consider n (1/2)r for increasing value of n i.e.
Solution:
n = 1 : 1. (1/2)1 = 1/2 = 0.5
n = 2 : 2 × (1/2)2 = 1/2 = 0.5
n = 3 : 3 × (1/2)3 = 0.375
n = 10 : 10 (1/2)10 = 0.00976, and so on
Thus we observe that as n → ∞
n rn → 0 for |r| < 1.
Illustration:
Evaluate 1 + 4/5 + 7/52 + 10/53 +…… to infinite terms.
Solution:
Let S = 1 + 4/5 + 7/52 + 10/53 + ………
1/5 S = 1/5 + 4/52 + 7/53 ………
Subtracting
(1–1/5) S = 1 + 3/5 + 3/52 + 3/53 + ………
4/5 S = 1/1–3/5 (? It is infinite G.P.)
⇒ S =25/8
Illustration:
Let t1, t2, t3, ……, tm–1, tm, tm+1, be a sequence so that
(i) tm+1/tm = tm/tm–1 ……… constant (r)
then tp = (t1)rp–1
(ii) tm+1/tm = tm/tm–1 = constant (r)
then tp = constant 1 + (constant 2) × rp–1
(iii) If the difference of difference of terms are in G.P. then
tp = a + bp + crp–1, where r is the common ratio.
Illustration:
7. 14. 33. 88. 251. 738 …………
Note:
324/108 = 108/36 = 36/12 = 3
∴ tp = a + bp + c 3p–1
p=1 t1 = 7 = a + b + c
p=2 t2 = 14 = a + 2b + 3c
p=3 t3 = 33 = a + 3b + 9c
Solving, we get a = 3, b = 1, c = 3
⇒ tp = 3 + p + 3. (3p–1)
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