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NCERT solution class 11 chapter 9 Sequences and Series exercise 9.4 mathematics

EXERCISE 9.4


Page No 196:

Question 1:

Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Answer:

The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

nth term, an = n ( n + 1)

Question 2:

Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Answer:

The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

nth term, an = n ( n + 1) ( n + 2)

= (n2 + n) (n + 2)

= n+ 3n+ 2n

Question 3:

Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …

Answer:

The given series is 3 ×12 + 5 × 22 + 7 × 32 + …

nth term, an = ( 2n + 1) n2 = 2n3 + n2

Question 3:

Find the sum to n terms of each of the series in Exercises 1 to 7.

3 × 12 + 5 × 22 + 7 × 32 + …

Answer:

The given series is 3 ×12 + 5 × 22 + 7 × 32 + …

nth term, an = ( 2n + 1) n2 = 2n3 + n2

Question 4:

Find the sum to n terms of the series 

Answer:

The given series is 

nth term, an = 

Adding the above terms column wise, we obtain

Question 5:

Find the sum to n terms of the series 

Answer:

The given series is 52 + 62 + 72 + … + 202

nth term, an = ( n + 4)2 = n2 + 8n + 16

16th term is (16 + 4)2 = 2022

Question 6:

Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Answer:

The given series is 3 × 8 + 6 × 11 + 9 × 14 + …

a= (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n2 + 15n

Question 7:

Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …

Answer:

The given series is 12 + (12 + 22) + (12 + 2+ 32 ) + …

an = (12 + 22 + 32 +…….+ n2)

= nn+12n+16=n2n2+3n+16=2n3+3n2+n6=13n3+12n2+16n

Question 8:

Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).

Answer:

an = n (n + 1) (n + 4) = n(n+ 5n + 4) = n3 + 5n2 + 4n

Question 9:

Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Answer:

an = n2 + 2n

Consider 

The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal to 2.

Therefore, from (1) and (2), we obtain

Question 10:

Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Answer:

an (2n – 1)2 = 4n2 – 4n + 1


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