Find the sum of n terms of the series, whose n th term is n 2 n 2 - 1 4 n 2 - 1

Given u n = n 2 n 2 - 1 4 n 2 - 1 = n 2 4 · 4 n 2 - 4 4 n 2 - 1 = n 2 4 · 4 n 2 - 1 - 3 4 n 2 - 1 = n 2 4 · 1 - 3 4 n 2 - 1 = n 2 4 - 3 n 2 4 · 4 n 2 - 1 = n 2 4 - 3 4 × 4 · 4 n 2 4 n 2 - 1 = n 2 4 - 3 16 4 n 2 - 1 + 1 4 n 2 - 1 = n 2 4 - 3 16 1 + 1 4 n 2 - 1 = n 2 4 - 3 16 - 3 16 4 n 2 - 1 ∴ S n = 1 4 ∑ n 2 - ∑ 3 16 - ∑ 3 16 4 n 2 - 1 ⇒ S n = 1 4 ∑ n 2 - 3 16 n - 3 16 ∑ 1 4 n 2 - 1 ⇒ S n = 1 4 ∑ n 2 - 3 16 n - 3 16 ∑ 1 2 n - 1 2 n + 1 ⇒ S n = 1 4 ∑ n 2 - 3 16 n - 3 16 S n " . . . i Now, we know that sum of series whose n th term is 1 u n = a + n b a + n + 1 b a + n + 2 b . . . . a + n + r - 1 b is given by S n = C - a + n b r - 1 b u n , where C is independent of n . ∴ S n " = C - 1 2 2 n + 1 Putting n = 1 , we have S 1 " = C - 1 2 · 3 ⇒ 1 1 · 3 = C - 1 2 · 3 ⇒ C = 1 2 So, S n " = 1 2 - 1 2 2 n + 1 Hence, by i , we have S n = 1 4 ∑ n 2 - 3 16 n - 3 16 1 2 - 1 2 2 n + 1 = n ( n + 1 ) ( 2 n + 1 ) 24 - 3 16 n - 3 16 1 2 - 1 2 ( 2 n + 1 ) = n ( n + 1 ) ( 2 n + 1 ) 24 - 3 16 n - 3 32 2 n 2 n + 1 = n ( n + 1 ) ( 2 n + 1 ) 24 - 3 16 4 n 2 + 4 n 2 ( 2 n + 1 ) = n ( n + 1 ) 8 2 n + 1 3 - 3 2 n + 1 = n ( n + 1 ) 4 n 2 + 4 n - 8 24 ( 2 n + 1 ) = n ( n + 1 ) ( n - 1 ) ( n + 2 ) 6 ( 2 n + 1 ) = ( n - 1 ) ( n ) ( n + 1 ) ( n + 2 ) 6 ( 2 n + 1 )

Find the sum of n terms of the series, whose n th term is n 4 + 2 n 3 + n 2 - 1 n 2 + n

In question the n th term is T n = n 4 + 2 n 3 + n 2 - 1 n 2 + n = n 2 × n 2 + 2 n + 1 - 1 n 2 + n = n 2 ( n + 1 ) 2 - 1 n ( n + 1 ) = n n + 1 - 1 n ( n + 1 ) Let the sum of the series is S n = Σ n n + 1 - 1 n ( n + 1 ) = Σ n n + 1 - Σ 1 n ( n + 1 ) . Let S n = S 1 - S 2 S 1 = Σ n 2 + n = n n + 1 2 n + 1 6 + n n + 1 2 = n n + 1 × 2 n + 1 + 3 6 = n n + 1 n + 2 3 S 2 = Σ 1 n n + 1 . = ∑ 1 n - 1 n + 1 = 1 - 1 2 + 1 2 - 1 3 + … + 1 n - 1 - 1 n + 1 n - 1 n + 1 = 1 - 1 n + 1 = n n + 1 ∴ S n = n ( n + 1 ) ( n + 2 ) 3 - n n + 1

Find the sum of n terms of the series whose n th term is n 3 + 3 n 2 + 2 n + 2 n 2 + 2 n

Given, u n = n 3 + 3 n 2 + 2 n + 2 n 2 + 2 n = n ( n + 1 ) ( n + 2 ) + 2 n ( n + 2 ) = ( n + 1 ) + 2 n ( n + 2 ) = ( n + 1 ) + 2 ( n + 1 ) n ( n + 1 ) ( n + 2 ) = ( n + 1 ) + 2 n + 2 n ( n + 1 ) ( n + 2 ) = ( n + 1 ) + 2 ( n + 1 ) ( n + 2 ) + 2 n ( n + 1 ) ( n + 2 ) ∴ S n = ∑ ( n + 1 ) + ∑ 2 ( n + 1 ) ( n + 2 ) + 2 n ( n + 1 ) ( n + 2 ) ∴ S n = n n + 1 2 + n + S n " . . . . i where, S n " = ∑ 2 ( n + 1 ) ( n + 2 ) + 2 n ( n + 1 ) ( n + 2 ) Now, we know that the sum of series u 1 + u 2 + u 3 + . . . . . + u n where 1 u n = a + n b a + n + 1 b a + n + 2 b . . . . a + n + r - 1 b is given by S n = C - a + n b r - 1 b u n , where C is independent of n . Therefore, ∴ S n " = C - 2 n + 2 - 1 ( n + 1 ) ( n + 2 ) By putting n = 1 , we get S 1 ' ' = C - 2 3 - 1 6 ⇒ 2 2 · 3 + 2 1 · 2 · 3 = C - 5 6 ⇒ C = 2 3 + 5 6 ⇒ C = 3 2 Hence, ∴ S n " = 3 2 - 2 n + 2 - 1 ( n + 1 ) ( n + 2 ) Therefore, by i , we have ∴ S n = 3 2 + n ( n + 1 ) 2 + n - 2 n + 2 - 1 ( n + 1 ) ( n + 2 ) = 3 2 + n ( n + 3 ) 2 - 2 n + 2 - 1 ( n + 1 ) ( n + 2 )

Find the sum of n terms of the series whose n th term is n 4 + n 2 + 1 n 4 + n

We have, u n = n 4 + n 2 + 1 n 4 + n = n 2 + n + 1 n 2 − n + 1 n n 3 + 1 = n 2 + n + 1 n 2 − n + 1 n n 2 − n + 1 ( n + 1 ) = n 2 + n + 1 n 2 + n = 1 + 1 n ( n + 1 ) ∴ S n = ∑ 1 + 1 n ( n + 1 ) = n + S n " . . . i where, S n " = 1 n n + 1 Now, we know that sum of series whose n th term is 1 u n = a + n b a + n + 1 b a + n + 2 b . . . . a + n + r - 1 b is given by S n = C - a + n b r - 1 b u n , where C is independent of n . Hence, S n " = C - 1 n + 1 Putting n = 1 , we get S 1 " = C - 1 2 ⇒ 1 1 · 2 = C - 1 2 ⇒ C = 1 Hence, S n " = 1 - 1 n + 1 Therefore, by i , we have ∴ S n = n + 1 - 1 ( n + 1 )

E M B I B E

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.

H S Hall and S R Knight Solutions for Chapter: Summation of Series, Exercise 1: EXAMPLES. XXIX. a.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Summation of Series, Exercise 1: EXAMPLES. XXIX. a.

Attempt the free practice questions on Chapter 29: Summation of Series, Exercise 1: EXAMPLES. XXIX. a. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.

Questions from H S Hall and S R Knight Solutions for Chapter: Summation of Series, Exercise 1: EXAMPLES. XXIX. a. with Hints & Solutions

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 16

Find the sum of $n$ terms of the series whose $n$ th term is $(n^{2} + 5 n + 4) (n^{2} + 5 n + 8)$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 17

Find the sum of $n$ terms of the series, whose $n^{th}$ term is $\frac{n^{2} (n^{2} - 1)}{4 n^{2} - 1}$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 18

Find the sum of $n$ terms of the series, whose $n^{th}$ term is $\frac{n^{4} + 2 n^{3} + n^{2} - 1}{n^{2} + n}$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 19

Find the sum of $n$ terms of the series whose $n th$ term is

$\frac{n^{3} + 3 n^{2} + 2 n + 2}{n^{2} + 2 n}$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 20

Find the sum of $n$ terms of the series whose $n^{th}$ term is $\frac{n^{4} + n^{2} + 1}{n^{4} + n}$

EASY

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 21

Show that the $n^{t h}$ term of the $r^{th}$ order of the figurate number is equal to the $r^{th}$ term of the $n^{t h}$ order.

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 22

If the $n^{th}$ term of the $r^{th}$ order of figurate number is equal to the $(n + 2)^{th}$ term of the $(r - 2)^{th}$ order; show that $r = n + 2$ .

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 29 - Summation of Series>EXAMPLES. XXIX. a.>Q 23

Show that the sum of the first $n$ of all the sets of polygonal numbers form the linear to that of the $r^{th}$ order inclusive is $\frac{(r - 1) n (n + 1)}{12} (r n - 2 n - r + 8)$ .

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