Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.6
Question 1
Maharashtra Board Solution
Find the sum .
Solution & Step-by-Step Answer:
Question 2
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Find
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
Find
Solution & Step-by-Step Answer:
= [(n + 1) + 2] = (n + 3)
Question 4
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Find
Solution & Step-by-Step Answer:
Question 5
Maharashtra Board Solution
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms.
Solution & Step-by-Step Answer:
5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms Now, 5, 9, 13, … are in A.P. with rth term = 5 + (r – 1) (4) = 4r + 1 7, 11, 15, ….. are in A.P. with rth term = 7 + (r – 1) (4) = 4r + 3 ∴ 5 × 7 + 9 × 11 + 13 × 15 + …… upto n terms
Question 6
Maharashtra Board Solution
Find the sum 22 + 42 + 62 + 82 + ….. upto n terms.
Solution & Step-by-Step Answer:
Question 7
Maharashtra Board Solution
Find (702 – 692) + (682 – 672) + (662 – 652) + …… + (22 – 12)
Solution & Step-by-Step Answer:
Let S = (702 – 692) + (682 – 672) + …… + (22 – 12) ∴ S = (22 – 12) + (42 – 32) + ….. + (702 – 692) Here, 2, 4, 6,…, 70 are in A.P. with rth term = 2r and 1, 3, 5, …,69 are in A.P. with rth term = 2r – 1
Question 8
Maharashtra Board Solution
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution & Step-by-Step Answer:
Let S = 1 × 3 × 5 + 3 × 5 × 7 + ….. upto n terms Here, 1, 3, 5, 7 … are in A.P. with rth term = 2r – 1, 3, 5, 7, 9,… are in A.P. with rth term = 2r + 1, 5, 7, 9, 11,… are in A.P. with rth term = 2r + 3
Question 9
Maharashtra Board Solution
If = , find n.
Solution & Step-by-Step Answer:
Question 10
Maharashtra Board Solution
If S1, S2 and S3 are the sums of first n natural numbers, their squares and their cubes respectively, then show that 9 = S3(1 + 8S1).
Solution & Step-by-Step Answer:
