Verified Solution • 2025–26 Session moderate

QJEEM25Ma3ea4: JEE Main 2025 Mathematics - Class 12 JEEM25Ma3ea4 Solution: Let be the values of , for which the equations , and have…

Let α,β (α≠β)\alpha, \beta \ (\alpha \neq \beta) be the values of mm , for which the equations x+y+z=1x+y+z=1 , x+2y+4z=mx+2y+4z=m and x+4y+10z=m2x+4y+10z=m^2 have infinitely many solutions. Then the value of ∑n=110(nα+nβ)\sum\limits_{n=1}^{10} (n^{\alpha}+n^{\beta}) is equal to :

Multiple Choice Options

  • A 3410
  • B 560
  • C 3080
  • D 440

Step-by-Step Verified Solution

amp;Δamp;=∣1amp;1amp;11amp;2amp;41amp;4amp;10∣=1(20−16)−1(10−4)+1(4−2)amp;=4−6+2=0amp; For infinite solutions amp;amp;Δx=Δy=Δz=0amp; m2−3x+2=0amp; m=1,2amp;α=1,β=2amp;∴∑n=110(nα+nβ)=∑n=110n1+∑n=110n2amp;=10(11)2+10(11)(21)6amp;=55+385amp;=440\begin{aligned}&\begin{aligned}\Delta & =\left|\begin{array}{ccc}1 & 1 & 1 \\1 & 2 & 4 \\1 & 4 & 10\end{array}\right|=1(20-16)-1(10-4)+1(4-2) \\& =4-6+2=0\end{aligned}\\&\text { For infinite solutions }\\&\begin{aligned}& \Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \\& \mathrm{~m}^2-3 \mathrm{x}+2=0 \\& \mathrm{~m}=1,2 \\& \alpha=1, \beta=2 \\& \therefore \sum_{\mathrm{n}=1}^{10}\left(\mathrm{n}^\alpha+\mathrm{n}^\beta\right)=\sum_{\mathrm{n}=1}^{10} \mathrm{n}^1+\sum_{\mathrm{n}=1}^{10} \mathrm{n}^2 \\&= \frac{10(11)}{2}+\frac{10(11)(21)}{6} \\&= 55+385 \\&= 440\end{aligned}\end{aligned}

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