Sum of series S - r -0 n 3 r -4- n C r - r -4 C 4- r -033 r - n -4 C r - n -4 C 4A- 4n-4r-0-n-3 Cn-n-3 Cn-1B- 4n-42 n C 4C- 4 n -4- n -4 m -4 C 4D- 3 n -4-2 n -4- n -4 C 4

The correct options are A 4^n+4^n+3C_n+ ^n+3C_n 1 C 4^n+4 ^n+4C_4^nC_r^r+4C_4=n!r!·(n r)!×4! r!(r+4)! =4!(n+1)(n+2)(n+3)(n+4)× (^n+4C_r+4) =^n+4C_r+4^n+4C_4 ...

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Byju's Answer

Standard XII

Mathematics

Integration

Sum of series...

Question

Sum of series $S = n \sum r = 0 \frac{3^{r + 4} \cdot^{n} C_{r}}{^{r + 4} C_{4}} + 3 \sum r = 0 \frac{3^{r} \cdot^{n + 4} C_{r}}{^{n + 4} C_{4}}$

\frac{4^{n + 4}}{^{n + 3} C_{n} +^{n + 3} C_{n - 1}}

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4^{n + 4} (^{2 n} C_{4})

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\frac{4^{n + 4}}{^{n + 4} C_{4}}

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\frac{3^{n + 4} + 2^{n + 4}}{^{n + 4} C_{4}}

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Solution

The correct options are
A $\frac{4^{n + 4}}{^{n + 3} C_{n} +^{n + 3} C_{n - 1}}$
C $\frac{4^{n + 4}}{^{n + 4} C_{4}}$
$\frac{^{n} C_{r}}{^{r + 4} C_{4}} = \frac{n!}{r! \cdot (n - r)!} \times \frac{4! r!}{(r + 4)!} = \frac{4!}{(n + 1) (n + 2) (n + 3) (n + 4)} \times (^{n + 4} C_{r + 4}) = \frac{^{n + 4} C_{r + 4}}{^{n + 4} C_{4}}$

So,
$S = \frac{1}{^{n + 4} C_{4}} [n \sum r = 0 3^{r + 4} \times^{n + 4} C_{4} + 3 \sum r = 0^{n + 4} C_{r} \cdot 3^{r}] = \frac{1}{^{n + 4} C_{4}} n + 4 \sum k = 0^{n + 4} C_{k} \cdot 3^{k} = \frac{4^{n + 4}}{^{n + 4} C_{4}}$

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Similar questions

Q. Sum of the series

S = \sum_{r = 0}^{n} \frac{3^{r + 4} ({^{n} C}_{r})}{{^{r + 4} C}_{4}} + \sum_{r = 0}^{3} \frac{{^{n + 4} C}_{r} 3^{r}}{{^{n + 4} C}_{4}}

Q. If

\frac{{^{n} C}_{r} + 4 {^{n} C}_{r + 1} + 6 {^{n} C}_{r + 2} + 4 {^{n} C}_{r + 3} + {^{n} C}_{r + 4}}{{^{n} C}_{r} + 3 {^{n} C}_{r + 1} + 3 {^{n} C}_{r + 2} + {^{n} C}_{r + 3}} = \frac{n + k}{r + k}

. Find the value of k

Q. If n is positive integer, then

∣ ∣ ∣ ∣ ∣ \begin{matrix} {n + 2}_{C_{n}} & n + 3_{C_{n + 1}} & n + 4_{C_{n + 2}} n + 3_{C_{n + 1}} & n + 4_{C_{n + 2}} & n + 5_{C_{n + 3}} n + 4_{C_{n + 2}} & n + 5_{C_{n + 3}} & n + 6_{C_{n + 4}} \end{matrix} ∣ ∣ ∣ ∣ ∣

is equal to

_{95} A m^{241} a n d_{90} T h^{234}

belong respectively to:

Q. Find the sum of

n

terms of the series whose

n^{t h}

term is

3 (4^{n} + 2 n^{2}) - {4 n}^{3}

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Sum of series S=n∑r=03r+4⋅ nCrr+4C4+3∑r=03r⋅ n+4Crn+4C4

The correct options are A 4n+4n+3Cn+ n+3Cn−1 C 4n+4 n+4C4nCrr+4C4=n!r!⋅(n−r)!×4! r!(r+4)!=4!(n+1)(n+2)(n+3)(n+4)×(n+4Cr+4)=n+4Cr+4n+4C4 So, S=1n+4C4[n∑r=03r+4× n+4C4+3∑r=0 n+4Cr⋅3r]=1n+4C4n+4∑k=0 n+4Ck⋅3k=4n+4n+4C4

Sum of series $S = n \sum r = 0 \frac{3^{r + 4} \cdot^{n} C_{r}}{^{r + 4} C_{4}} + 3 \sum r = 0 \frac{3^{r} \cdot^{n + 4} C_{r}}{^{n + 4} C_{4}}$