Let - Zr-r- - r- r-1-2-3-n- Then - -r-1n Zr- is less than

The correct option is C n(n+1) |Z_r r| ≤ r ∴ Z_r ≤ 2r or |Z_r| ≤ 2r also ∑_r=1^n |Z_r| ≤ 2 ∑_r=1^n r also |∑_r=1^n Z_r| ≤∑_r=1^ ...

You visited us 5 times! Enjoying our articles? Unlock Full Access!

Properties of Modulus

Let | Zr-r|...

Question

Let $| Z_{r} - r | \leq r, \forall r = 1, 2, 3, . . . . . . . n .$ Then $∣ ∣ ∣ ∣ n \sum r = 1 Z_{r} ∣ ∣ ∣ ∣$ is less than

n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n (n + 1)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{n (n + 1)}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

| z_{r} - r | \leq r

r = 1, 2, 3, . . ., n .

∣ ∣ \sum_{r = 1}^{n} z_{r} ∣ ∣

n

r

n^{r} - n {(n - 1)}^{r} + \frac{n (n - 1)}{2!} {(n - 2)}^{r} - \frac{n (n - 1) (n - 2)}{3!} {(n - 3)}^{r} + \dots

0

r

n

n!

r = n

$r^{t h}$ term in the expansion of $(a + 2 x)^{n}$ is

\frac{1}{m!} C_{0} + \frac{n}{(m + 1)!} C_{1} + \frac{n (n - 1)}{(m + 2)!} C_{2} + . . . . . + \frac{n (n - 1) . . .2 .1}{(m + n)!} C_{n} = \frac{(m + n + 1) (m + n + 2) . . . (m + 2 n)}{(m + n)!}

r \geq 0

(\begin{matrix} m r \end{matrix}) C_{0} + (\begin{matrix} m r - 1 \end{matrix}) C_{1} + . . . . . . (\begin{matrix} m 0 \end{matrix}) C_{r} = (\begin{matrix} m + n r \end{matrix})

Join BYJU'S Learning Program