OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Ex 14(h)

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S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(h)

Question 1.
1 + 2x + 3x² + 4x³ + ………
Solution:
Let S_n = 1 + 2x + 3x² + 4x³ + ……… n terms
i.e. S_n = 1 + 2x + 3x² + 4x³ + \ldots \ldots \ldots \ldots . .+(n-1) x^{n-2}+n x^{n-1}$
It is an arithmetico-geometric series.
x S_n = x + 2x² + 3x³ + ………. + (n + 1) x^{n – 1} + nxⁿ
eqn. (1) – eqn. (2) gives;
(1 – x)S_n = 1 + x + x² + ……. + x^{n – 1} – nxⁿ = \(\frac{1\left(1-x^n\right)}{1-x}\) – nxⁿ
⇒ S_n = \(\frac{1-x^n}{(1-x)^2}\) – \(\frac{n x^n}{1-x}\)

Question 2.
1 + 3x +5x² + 7x³ + …..
Solution:
Let S_n = 1 + 3x + 5x² + 7x³ + …… +(2n – 3) x^n-2 + (2n – 1) x^n-1 …(1)
It is an A.G.P.
∴ xS_n = x + 3x² + 5x³ + ….. +(2n – 3) x^n-1 + (2n – 1) xⁿ …(2)
eqn. (1) – eqn. (2) gives ;
(1 – x)S_n = 1 + 2x + 2x² + 2x³ – …. + 2x^n-1 – (2n – 1) xⁿ
= 1 + \(\frac{2 x\left[1-x^{n-1}\right]}{1-x}\) – (2n – 1)xⁿ = 1 + \(\frac{2 x}{1-x}\) – \(\frac{2 x^n}{1-x}\) – (2n – 1)xⁿ
∴ S_n = \(\frac{1+x}{(1-x)^2}\) – \(\frac{2 x^n}{(1-x)^2}\) – \(\frac{(2 n-1) x^n}{1-x}\)

Question 3.
2.1 + 3.2 + 4.4 + 5.8 + ….
Solution:
Let S_n = 2.1 + 3.2 + 4.4 + 5.8 + …… n 2^n-2+(n+1) 2^n-1 …(1)
∴ 2S_n = 2.2 + 3.4 + 4.8 + …….. + n 2^n-1 + (n+1) 2ⁿ …(2)
eqn. (1) – eqn. (2) gives ;
-S_n = 2.1 + 1.2 + 1.4 + 1.8 ……. + 2^n-1 – (n + 1) 2ⁿ = 2 + \(\frac{2\left(2^{n-1}-1\right)}{2-1}\) – (n + 1)2ⁿ
⇒ -S_n = 2 + 2 (2^n-1 – 1) – (n + 1) 2ⁿ = 2 + 2ⁿ – 2 – n2ⁿ – 2ⁿ = -n2ⁿ
⇒ S_n = n.2ⁿ

Question 4.
\(\frac{1}{2}\) + \(\frac{3}{6}\) + \(\frac{5}{18}\) + ……
Solution:
Let S = \(\frac{1}{2}\) + \(\frac{3}{6}\) + \(\frac{5}{18}\) + …… ∞ …(1)
It is an arithmetico geometric series.

Question 5.
\(\frac{3}{2}\) – \(\frac{5}{6}\) + \(\frac{7}{18}\) – ……
Solution:
Let S = \(\frac{3}{2}\) – \(\frac{5}{6}\) + \(\frac{7}{18}\) – …… ∞ …(1)
It is an arithmetico geometric series with r = \(\frac{-1}{3}\)
∴ \(\frac{-1}{3} S\) = \(\frac{-3}{6}\) + \(\frac{5}{18}\) + ….. ∞ …(2)
eqn. (1) – eqn. (2) gives ;
\(\frac{4}{3} S\) = \(\frac{3}{2}\) – \(\frac{2}{6}\) + \(\frac{2}{18}\) + ….. ∞
= \(\frac{3}{2}\) + \(\frac{-\frac{2}{6}}{1+\frac{1}{3}}\) = \(\frac{3}{2}\) + \(\left(-\frac{1}{3}\right)\) × \(\frac{3}{4}\) = \(\frac{3}{2}\) – \(\frac{1}{4}\) = \(\frac{5}{4}\) ⇒ S = \(\frac{5}{4}\) × \(\frac{3}{4}\) = \(\frac{15}{16}\)

Question 6.
1 – \(\frac{2}{5}\) + \(\frac{3}{5^2}\) – \(\frac{4}{5^3}\) + …..
Solution:
Let S = 1 – \(\frac{2}{5}\) + \(\frac{3}{5^2}\) – \(\frac{4}{5^3}\) + ….. ∞ …(1)
∴ – \(\frac{1}{5}\)S = – \(\frac{1}{5}\) + \(\frac{2}{5^2}\) – \(\frac{3}{5^3}\) + ….. ∞ …(2)
eqn. (1) – eqn. (2) gives ;
\(\frac{6}{5}\)S = 1 – \(\frac{1}{5}\) + \(\frac{1}{5^2}\) …… ∞
= \(\frac{1}{1+\frac{1}{5}}\)

⇒ \(\frac{6}{5}\)S = \(\frac{5}{6}\)
⇒ S = \(\frac{25}{36}\)

Question 7.
Sum up the series \(\frac{2}{3}\) + \(\frac{5}{9}\) + \(\frac{8}{27}\) + \(\frac{11}{81}\) + …. to n terms and hence find the sum to infinity.
Solution:

Question 8.
1 + 4x² + 7x⁴ + ……
Solution:
Let S = 1 + 4x² + 7x⁴ + ……… ∞ …(1)
x²S = x² + 4x⁴ + ……… ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(1 – x)² S = 1 + 3x² + 3x⁴ + …….. ∞
= 1 + \(\frac{3 x^2}{1-x^2}\) = \(\frac{1-x^2+3 x^2}{1-x^2}\) = \(\frac{1+2 x^2}{\left(1-x^2\right)}\)
⇒ S = \(\frac{1+2 x^2}{\left(1-x^2\right)^2}\)

Question 9.
1 – x + 2x² – 3x³ + 4x⁴ – …………
Solution:
Let S = 1 – x + 2x² – 3x³ + …… ∞
⇒ S = 1 – {x – 2x² + 3x³ + …… ∞} …(1)
which is an A.G.P with common ratio (-x).
∴ -xS = -x + x² – 2x³ + …… ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(1 + x)S = 1 + x² – x³ – …… ∞
= 1 + \(\frac{x^2}{1+x}\)
⇒ S = \(\frac{1}{1+x}\) + \(\frac{x^2}{(1+x)^2}\) = \(\frac{1+x+x^2}{(1+x)^2}\)

Question 10.
1² + 3²x + 5²x² + 7²x³ + …..
Solution:
Let S = 1² + 3²x + 5²x² + 7²x³ + …… ∞
∴ S = 1 + 9x + 25x² + 49x³ + …… ∞ …(1)
xS = x + 9x² + 25x³ + …… ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(1 – x) S = 1 + 8x + 16x² + 24x³ + …… ∞ …(3)
It is clearly an A.G.P series.
∴ x(1 – x)S = x + 8x² + 16x³ + …… ∞ …(4)
eqn. (3) – eqn. (4) gives;
(1 – x)²S = 1 + 7x + 8x² + 8x³ + …… ∞
= 1 + 7x + \(\frac{8 x^2}{1-x}\)
⇒ S = \(\frac{1+7 x}{(1-x)^2}\) + \(\frac{8 x^2}{(1-x)^3}\)

Question 11.
Show that the square root of \(3^{\frac{1}{2}} \times 9^{\frac{1}{4}} \times 27^{\frac{1}{8}} \times 81^{\frac{1}{16}}\) × ……. to infinity is 3 .
[Note. When we are asked to find the square root of a number, it is presumed that we have to find the principal square root.]
Solution: