The sum S n - k -0 n -1 K -3 n C k- where n -12- - isA- None of these B- -1n-3 n-1 Cn-1C- -1n-3 n-1 CnD- -1n-3 n-1 Cn-1

The correct option is C S_n=^3nC_0 ^3nC_1 + ^3nC_2 +..... + ( 1)^n. ^3nC_n But ^3nC_0 = ^3n 1C_0 ^3nC_1 = ^3n 1C_0 ^ 3n 1C_1 ^3nC_2 = ^3n 1C_1 +^3n 1 ...

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

Byju's Answer

Standard XI

Mathematics

A.G.P

The sum S n =...

Question

The sum $S_{n} = \sum_{k = 0}^{n} (- 1)^{K} .^{3 n} C_{k}, w h e r e n = 12, . . . . . .$ is

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

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

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

None of these

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

Open in App

Solution

The correct option is B
$S_{n} =^{3 n} C_{0} -^{3 n} C_{1} +^{3 n} C_{2} + . . . . . + (- 1)^{n} .^{3 n} C_{n} B u t^{3 n} C_{0} =^{3 n - 1} C_{0} -^{3 n} C_{1} = -^{3 n - 1} C_{0} -^{3 n - 1} C_{1}^{3 n} C_{2} =^{3 n - 1} C_{1} +^{3 n - 1} C_{2} -^{3 n} C_{3} = -^{3 n - 1} C_{2} -^{3 n - 1} C_{3} . . . . . . . . . . . . . . . . . . . . . . . . . . . . (- 1)^{n} .^{3 n} C_{n} = (- 1)^{n} .^{3 n - 1} C_{1} + (- 1)^{n} .^{3 n - 1} C_{n}$
On adding we get $S_{n} = (- 1)^{n} .^{3 n - 1} C_{n}$

Suggest Corrections

Similar questions

Q. The sum

S_{n} = \sum_{k = 0}^{n} (- 1)^{K} .^{3 n} C_{k},

Q. The value of

\frac{1}{2}^{n} C_{1} .^{n - 1} C_{1} + \frac{2}{3}^{n} C_{2} + \frac{3}{4}^{n} C_{c} .^{n - 1} C_{3} + \dots \dots

Q. If

(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{n},

then

\frac{C_{0}}{1 \cdot 2} 2^{2} + \frac{C_{1}}{2 \cdot 3} 2^{3} + \frac{C_{2}}{3 \cdot 4} 2^{4} + \dots + \frac{C_{n}}{(n + 1) (n + 2)} 2^{n + 2}

is equal to

Q. The

n^{th}

term in the expansion of

{log}_{e} (\frac{4}{3})

$\sum_{r = 1}^{n} (\sum_{k = 0}^{r - 1}^{n} C_{r}^{r} C_{k} 2^{k})$ is equal to

Join BYJU'S Learning Program

The sum Sn=∑nk=0(−1)K.3nCk,where n=12,...... is

The correct option is B Sn=3nC0−3nC1+3nC2+.....+(−1)n.3nCnBut3nC0=3n−1C0−3nC1=−3n−1C0−3n−1C13nC2=3n−1C1+3n−1C2−3nC3=−3n−1C2−3n−1C3............................(−1)n.3nCn=(−1)n.3n−1C1+(−1)n.3n−1Cn On adding we get Sn=(−1)n.3n−1Cn

The sum $S_{n} = \sum_{k = 0}^{n} (- 1)^{K} .^{3 n} C_{k}, w h e r e n = 12, . . . . . .$ is