The sum is 3266+r=0∑24325−r10⋅2r
=3266+10⋅3251r=0∑24(2⋅3)r=3266+32510⋅5625−1
=3266+2⋅3(625−1)=3266+2⋅625⋅3−6=3266⋅625
=326626=(36)26=226
JEE Main 2026 — Mathematics Algebra
3266+32510⋅1+32410⋅2+32310⋅22+…+310⋅224 is equal to :
Held on 28 Jan 2026 · Verified 6 Jul 2026.
226
325
326
225
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If $\alpha=1$ and $\beta=1+i\sqrt{2}$, where $i=\sqrt{-1}$ are two roots of the equation $x^3+ax^2+bx+c=0$, $a,b,c \in \mathbb{R}$, then $\int_{-1}^{1}(x^3+ax^2+bx+c)dx$ is equal to:
Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $x^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}, 0 \leq \mathrm{r} \leq \mathrm{n}$. If $P_{n}=C_{0}-C_{1}+\frac{2^{2}}{3} C_{2}-\frac{2^{3}}{4} C_{3}+\ldots. .+\frac{(-2)^{n}}{n+1} C_{n}$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2 n}}$ equals.
Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$, $z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:
Three persons enter in a lift at the ground floor. The lift will go upto $10^{\text {th }}$ floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to $\_\_\_\_$.
Work through every JEE Main Algebra PYQ, year by year.