11th Commerce Maths 1 Chapter 4 Exercise 4.5 Answers Maharashtra Board

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.5 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.5 Questions and Answers.

Std 11 Maths 1 Exercise 4.5 Solutions Commerce Maths

Question 1.
Find the sum \(\sum_{r=1}^{n}(r+1)(2 r-1)\).
Solution:

Question 2.
Find \(\sum_{r=1}^{n}\left(3 r^{2}-2 r+1\right)\).
Solution:

Question 3.
Find \(\sum_{r=1}^{n} \frac{1+2+3+\ldots+r}{r}\).
Solution:

Question 4.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+\ldots+r^{3}}{r(r+1)}\).
Solution:
We know that,

Question 5.
Find the sum 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms.
Solution:
5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + ….. upto n terms
Now, 5, 7, 9, 11, … are in A.P.
rth term = 5 + (r – 1) (2) = 2r + 3
7, 9, 11,. … are in A.P.
rth term = 7 + (r – 1) (2) = 2r + 5
∴ 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms

Question 6.
Find the sum 2 ² + 4 ² + 6 ² + 8 ² + …… upto n terms.
Solution:
2 ² + 4 ² + 6 ² + 8 ² + …… upto n terms
= (2 × 1) ² + (2 × 2) ² + (2 × 3) ² + (2 × 4) ² + ……

Question 7.
Find (70 ² – 69 ² ) + (68 ² – 67 ² ) + (66 ² – 65 ² ) + ……. + (2 ² – 1 ² )
Solution:
Let S = (70 ² – 69 ² ) + (68 ² – 67 ² ) + …… +(2 ² – 1 ² )
∴ S = (2 ² – 1 ² ) + (4 ² – 3 ² ) + …… + (70 ² – 69 ² )
Here, 2, 4, 6,…, 70 is an A.P. with rth term = 2r
and 1, 3, 5,….., 69 in A.P. with rth term = 2r – 1

Question 8.
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution:
1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Now, 1, 3, 5, 7, … are in A.P. with a = 1 and d = 2.
∴ rth term = 1 + (r – 1)2 = 2r – 1
3, 5, 7, 9, … are in A.P. with a = 3 and d = 2
∴ rth term = 3 + (r – 1)2 = 2r + 1
and 5, 7, 9, 11, … are in A.P. with a = 5 and d = 2
∴ rth term = 5 + (r – 1)2 = 2r + 3
∴ 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… upto n terms

= n(n + 1)[2n(n + 1) + 4n + 2 – 1] – 3n
= n(n + l)(2n ² + 6n + 1) – 3n
= n(2n ³ + 8n ² + 7n + 1 – 3)
= n(2n ³ + 8n ² + 7n – 2)

Question 9.
Find n, if \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots+\text { upto } n \text { terms }}{1+2+3+4+\ldots+\text { upto } n \text { terms }}\) = \(\frac{100}{3}\)
Solution:

Question 10.
If S ₁ , S ₂ , and S ₃ are the sums of first n natural numbers, their squares, and their cubes respectively, then show that:
9\(S_{2}^{2}\) = S ₃ (1 + 8S ₁ ).
Solution: