JEE Main 2024 — Mathematics Algebra

Given,

$$\alpha ={1}^{2}+{4}^{2}+{8}^{2}\ldots$$α=12+42+82…

Now, let $\sqrt{{t}_{n}}=a{n}^{2}+bn+c$tn​​=an2+bn+c

$$1=a+b+c....(i)$$1=a+b+c....(i)

$$4=4a+2b+c....(ii)$$4=4a+2b+c....(ii)

$$8=9a+3b+c....(iii)$$8=9a+3b+c....(iii)

Now, solving equation (i),(ii)&(iii) we get,

$$a=\frac{1}{2},b=\frac{3}{2},c=-1$$a=21​,b=23​,c=−1

Hence, ${t}_{n}={(\frac{{n}^{2}}{2}+\frac{3n}{2}-1)}^{2}$tn​=(2n2​+23n​−1)2

$$\alpha =\sum _{n=1}^{10}{(\frac{{n}^{2}}{2}+\frac{3n}{2}-1)}^{2}$$α=n=1∑10​(2n2​+23n​−1)2

$\displaystyle \Rightarrow 4\alpha =\sum _{n=1}^{10}{({n}^{2}+3n-2)}^{2}$⇒4α=n=1∑10​(n2+3n−2)2 and $\displaystyle \beta =\sum _{n=1}^{10}{n}^{4}$β=n=1∑10​n4

$$\Rightarrow 4\alpha -\beta =\sum _{n=1}^{10}(6{n}^{3}+5{n}^{2}-12n+4)$$⇒4α−β=n=1∑10​(6n3+5n2−12n+4)

$$\Rightarrow 4\alpha -\beta =(6\sum _{n=1}^{10}{n}^{3}+5\sum _{n=1}^{10}{n}^{2}-12\sum _{n=1}^{10}n+4\sum _{n=1}^{10}1)$$⇒4α−β=(6n=1∑10​n3+5n=1∑10​n2−12n=1∑10​n+4n=1∑10​1)

$$\Rightarrow 4\alpha -\beta =(6{(\frac{10(10+1)}{2})}^{2}+5(\frac{10\cdot 11\cdot 21}{6})-12(\frac{10\cdot 11}{2})+4\cdot 10)$$⇒4α−β=(6(210(10+1)​)2+5(610⋅11⋅21​)−12(210⋅11​)+4⋅10)

$$\Rightarrow 4\alpha -\beta =(6{(55)}^{2}+5\cdot 385-12\cdot 55+40)$$⇒4α−β=(6(55)2+5⋅385−12⋅55+40)

$$\Rightarrow 4\alpha -\beta =55(353)+40$$⇒4α−β=55(353)+40

Hence, on comparing with $55k+40$55k+40 we get, $k=353$k=353