Let an=3n+n2−1 and bn=2(n2−n+n2+n),n≥1 , then the value of a1−b1+a2−b2+…+a49−b49=c+d2 for some integer c and d then which is/are INCORRECT ?

d − 2 c = 14 , c + 2 d = 3 , c + d = 1

a n − b n = − 1 2 [ ( n + 1 − n ) − ( n − n − 1 ) ] = 4 − 5 2

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Let $a_{n} = 3 n + \sqrt{n^{2} - 1}$ and $b_{n} = 2 (\sqrt{n^{2} - n} + \sqrt{n^{2} + n}), n \geq 1$ , then the value of $\sqrt{a_{1} - b_{1}} + \sqrt{a_{2} - b_{2}} + \dots + \sqrt{a_{49} - b_{49}} = c + d \sqrt{2}$ for some integer $c$ and $d$ then which is/are INCORRECT ?

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$d - 2 c = 14$

$c + 2 d = 3$

$d - c = 9$

$c + d = 1$

answer is A, B, D.

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$\sqrt{a_{n} - b_{n}} = - \frac{1}{\sqrt{2}} [(\sqrt{n + 1} - \sqrt{n}) - (\sqrt{n} - \sqrt{n - 1})] = 4 - 5 \sqrt{2}$

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