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

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.5

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: