MCQ
જો $1^3+2^3+3^3+......+n^3=2025$ હોઈ તો $n=.......$
- ✓9
- B1
- C3
- D2
ઉકેલ : અહી $1^3+2^3+3^3+....+n^3=2025$
$\therefore \sum n^3=2025$
$\therefore \frac{n^2(n+1)^2}{4}=2025$
$\therefore \frac{n(n+1)}{2}=45$
$\therefore n^2+n-90=0.$ આથી $(n+10)(n-9)=0$ આથી$n=9$
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