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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The sum to $20$ terms of the series $2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} - \dots$ is equal to

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Explanation

Solution:

Hamein series ke 20 terms ka sum nikalna hai:

$S_{20} = 2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} - 7^{2} + \dots \text{ (upto 20 terms)}$

Step 1: Series ko do alag parts mein divide karein

Is series mein 10 positive terms (even numbers ke square) aur 10 negative terms (odd numbers ke square) hain.

$S_{20} = (2 \cdot 2^{2} + 2 \cdot 4^{2} + 2 \cdot 6^{2} + \dots + 2 \cdot 20^{2}) - (3^{2} + 5^{2} + 7^{2} + \dots + 21^{2})$

Step 2: Pattern ko simplify karein

Ise likhne ka ek asan tarika yeh hai ki hum har pair ko solve karein:

$S_{20} = (2^{2} + 2^{2} - 3^{2}) + (4^{2} + 4^{2} - 5^{2}) + (6^{2} + 6^{2} - 7^{2}) + \dots + (20^{2} + 20^{2} - 21^{2})$

Ab identity $a^{2} - b^{2} = (a-b)(a+b)$ ka use karte hain:

$(2^{2} - 3^{2}) = (2-3)(2+3) = -1(5) = -5$

$(4^{2} - 5^{2}) = (4-5)(4+5) = -1(9) = -9$

$(20^{2} - 21^{2}) = (20-21)(20+21) = -1(41) = -41$

Toh hamari series ban jayegi:

$S_{20} = (2^{2} - 5) + (4^{2} - 9) + (6^{2} - 13) + \dots + (20^{2} - 41)$

Step 3: Summation form mein likhein

General term $T_{r}$ (jahan $r = 1$ to $10$ ):

$T_{r} = (2r)^{2} - (4r + 1)$

$S_{10} = \sum_{r=1}^{10} (4r^{2} - 4r - 1)$

Step 4: Formula lagayein

$\sum r^{2} = \frac{n(n+1)(2n+1)}{6}, \quad \sum r = \frac{n(n+1)}{2}$

Jahan $n = 10$ :

$4 \sum_{r=1}^{10} r^{2} = 4 \times \frac{10 \times 11 \times 21}{6} = 4 \times 385 = 1540$

$-4 \sum_{r=1}^{10} r = -4 \times \frac{10 \times 11}{2} = -220$

$\sum_{r=1}^{10} -1 = -10$

Total Sum:

$S_{20} = 1540 - 220 - 10 = 1310$

Sahi jawab 1310 hai.

JEE
If $a_{0}$ is the greatest term in the sequence

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