Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.6 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.6

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 \(\sum_{r=1}^{n}\left(\frac{1+2+3 \ldots .+r}{r}\right)\)

Solution:

= \(\frac{n}{4}\) [(n + 1) + 2]

= \(\frac{n}{4}\) (n + 3)

Question 4.

Find \(\sum_{r=1}^{n}\left(\frac{1^{3}+2^{3}+\ldots . .+r^{3}}{r(r+1)}\right)\)

Solution:

Question 5.

Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms.

Solution:

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.

Find the sum 2^{2} + 4^{2} + 6^{2} + 8^{2} + ….. upto n terms.

Solution:

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 are in A.P. with rth term = 2r

and 1, 3, 5, …,69 are 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:

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.

If \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } n \text { terms }}{1+2+3+4+\ldots \text { upto } n \text { terms }}\) = \(\frac{100}{3}\), find n.

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\(\mathrm{S}_{2}{ }^{2}\) = S_{3}(1 + 8S_{1}).

Solution: