Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

EXPLANATION.

⇒ 1 + (1 - 2².1) + (1 - 4².3) + (1 - 6².5) + . . . . . + (1 - 20².19) = α - 220β.

As we know that,

General terms of an ap.

⇒ Tₙ = a + (n - 1)d.

Using this formula in this question, we get.

2, 4, 6, . . . . . 20 are in ap.

First term : a = 2.

Common difference : d = b - a : 4 - 2 = 2.

⇒ Tₙ = 2 + (n - 1)2.

⇒ Tₙ = 2 + 2n - 2.

⇒ Tₙ = 2n.

Now, we find n terms of ap.

⇒ 20 = 2 + (n - 1)2.

⇒ 20 = 2 + 2n - 2.

⇒ 20 = 2n.

⇒ n = 10.

Number of terms : n = 10.

1, 3, 5, . . . . . 19 are in ap.

First term : a = 1.

Common difference : d = b - a : 3 - 1 = 2.

⇒ Tₙ = 1 + (n - 1)2.

⇒ Tₙ = 1 + 2n - 2.

⇒ Tₙ = 2n - 1.

Now, we find n terms of ap.

⇒ 19 = 1 + (n - 1)2.

⇒ 19 = 1 + 2n - 2.

⇒ 19 = 2n - 1.

⇒ 19 + 1 = 2n.

⇒ 20 = 2n.

⇒ n = 10.

Number of terms : n = 10.

Now, we can write whole equation as,

$\sf \displaystyle 1 + \sum_{n = 1}^{10} \bigg[ 1 - (2n)^{2} (2n - 1) \bigg]$

$\sf \displaystyle 1 + \sum_{n = 1}^{10} \bigg[ 1 - 4n^{2} (2n - 1) \bigg]$

$\sf \displaystyle 1 + \sum_{n = 1}^{10} \bigg[ 1 - 8n^{3} + 4n^{2} \bigg]$

$\sf \displaystyle 1 + \sum_{n = 1}^{10} 1 - \sum_{n = 1}^{10} 8n^{3} + \sum_{n = 1}^{10} 4n^{2}$

As we know that,

Some standard results.

Sum of squares of first n natural numbers.

$\sf \displaystyle \sum_{r = 1}^{n} r^{2} = \frac{n(n + 1)(2n + 1)}{6}$

Sum of cubes of first n natural numbers.

$\sf \displaystyle \sum_{r = 1}^{n} r^{3} = \bigg[\frac{n(n + 1)}{2} \bigg]^{2}$

Using this results in this question, we get.

$\sf \displaystyle 1 + \sum_{n = 1}^{10} n - 8 \sum_{n = 1}^{10} n^{3} + 4 \sum_{n = 1}^{10} n^{2}$

$\sf \displaystyle 1 + 10 - 8 \bigg[\frac{10(10 + 1)}{2} \bigg]^{2} + 4 \bigg[\frac{10 (10 + 1)(2 \times 10 + 1)}{6} \bigg]$

$\sf \displaystyle 11 - \frac{8}{4} (10 \times 11)^{2} + 4 \bigg[\frac{10 \times 11 \times 21}{6} \bigg]$

$\sf \displaystyle 11 - 2 (10 \times 11)^{2} + 4 (5 \times 11 \times 7 )$

$\sf \displaystyle 11 - 2(12100) + 4(385)$

$\sf \displaystyle 11 - 24200 + 1540$

$\sf \displaystyle 11 - 22660$

$\sf \displaystyle 11 - 220 \times (103)$

$\sf \displaystyle 11 - 220 \times (103) = \alpha - 220\beta$

$\sf \displaystyle \alpha = 11 \ \ \ and \ \ \ \beta = 103$

Option [2] is correct answer.