Sum of the series - k-0 r -1 r - n- r - is

The correct option is A ( 1) ^ r [ n 1; r ] To find, S = ^ n C _ 0 ^ n C _1 + ^ n C _ 2 ^ n C _ 3 + ^ n C _ 4 .........+ ( 1 ) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

Sum of the se...

Question

Sum of the series
$\sum_{k = 0}^{r} {(- 1)}^{r} (\begin{matrix} n r \end{matrix})$ is

{(- 1)}^{r} (\begin{matrix} n r \end{matrix})

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

{(- 1)}^{r} (\begin{matrix} n - 1 r \end{matrix})

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

0

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

1

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

Open in App

Solution

The correct option is A ${(- 1)}^{r} (\begin{matrix} n - 1 r \end{matrix})$
To find,
$S =^{n} C_{0} -^{n} C_{1} +^{n} C_{2} -^{n} C_{3} +^{n} C_{4} . . . . . . . . . + {(- 1)}^{r}^{n} C_{r}$
We know that, $^{n} C_{r} =^{n - 1} C_{r - 1} +^{n - 1} C_{r}$ .
$∴$ $S =^{n} C_{0} - (^{n - 1} C_{0} +^{n - 1} C_{1}) + (^{n - 1} C_{1} +^{n - 1} C_{2}) -^{n - 1} C_{2} +^{n - 1} C_{3}) + . . . . . {(- 1)}^{r} (^{n - 1} C_{r - 1} +^{n - 1} C_{r}$ .)
$=^{n} C_{0} -^{n - 1} C_{0} -^{n - 1} C_{1} +^{n - 1} C_{1} +^{n - 1} C_{2} -^{n - 1} C_{2} + . . . . . {(- 1)}^{r} (^{n - 1} C_{r - 1} +^{n - 1} C_{r}$ .)
$S = {(- 1)}^{r}^{n - 1} C_{r}$
Hence the answer is ${(- 1)}^{r}^{n - 1} C_{r} .$

Suggest Corrections

Similar questions

Which of the following best describes the terms of the series? $\frac{1}{2}$ $^{n} C_{1}$ - $\frac{2}{3}$ $^{n} C_{2}$ + $\frac{3}{4}$ $^{n} C_{3}$ ................... $(- 1)^{n + 1}$ $\frac{1}{n + 1}$ $^{n} C_{n}$

(\begin{matrix} n + 1 r + 1 \end{matrix}) : (\begin{matrix} n r \end{matrix}) : (\begin{matrix} n - 1 r - 1 \end{matrix}) = 11 : 6 : 3,,

find

n

r

Q. Sum of series

\sum_{r = 1}^{n} (r^{2} + 1) r!

Q. The sum of the series

n \sum r = 1 (- 1)^{r - 1} .^{n} C_{r} (a - r)

Q. Sum the following series to n terms is

\sum_{r = 1}^{n} r (r + 1) (r + 2) (r + 3)

Join BYJU'S Learning Program

Sum of the series∑rk=0(−1)r(nr) is

Sum of the series
$\sum_{k = 0}^{r} {(- 1)}^{r} (\begin{matrix} n r \end{matrix})$ is