If S n - r -1 n - t -0 r -11-6 n n C r r C t 4 t - then the value of l- where l - n -1-1- S n is

S_n= ∑_r=1^n nbsp; ∑_t=0^r 1( 16^n ^nC_r ^rC_t 4^t) ⇒ S_n= ∑_r=1^n( 16^n ^nC_r ( ∑_t=0^r 1 ^rC_t 4^t)) = ∑_r=1^n( 16^n ^nC_r [(1+4)^r 4^r ]) nbsp; nbs ...

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

Byju's Answer

Standard XII

Mathematics

Binomial Expression

If S n =∑ r =...

Question

If $S_{n} = n \sum r = 1 r - 1 \sum t = 0 (\frac{1}{6^{n}}^{n} C_{r}^{r} C_{t} 4^{t}),$ then the value of $l,$ where $l = \infty \sum n = 1 (1 - S_{n})$ is

Open in App

Solution

$S_{n} = n \sum r = 1 r - 1 \sum t = 0 (\frac{1}{6^{n}}^{n} C_{r}^{r} C_{t} 4^{t})$
$\Rightarrow S_{n} = n \sum r = 1 (\frac{1}{6^{n}}^{n} C_{r} (r - 1 \sum t = 0^{r} C_{t} 4^{t}))$
$= n \sum r = 1 (\frac{1}{6^{n}}^{n} C_{r} [(1 + 4)^{r} - 4^{r}])$
(Using $(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{r})$
$= \frac{1}{6^{n}} n \sum r = 1 (^{n} C_{r} 5^{r} -^{n} C_{r} 4^{r})$
$= \frac{1}{6^{n}} [(1 + 5)^{n} - 1 - ((1 + 4)^{n} - 1)]$
$= \frac{6^{n} - 5^{n}}{6^{n}} = 1 - {(\frac{5}{6})}^{n}$

Now, $\infty \sum n = 1 (1 - S_{n}) = \infty \sum n = 1 {(\frac{5}{6})}^{n}$
$= \frac{5 / 6}{1 - 5 / 6} = 5$

Suggest Corrections

Similar questions

Q. If

S_{n} = n \sum r = 1 r - 1 \sum t = 0 (\frac{1}{6^{n}}^{n} C_{r}^{r} C_{t} 4^{t}),

then the value of

l

where

l = \infty \sum n = 1 (1 - S_{n})

Q. Let

S_{n} = \frac{n}{(n + 1) (n + 2)} + \frac{n}{(n + 2) (n + 4)} + \frac{n}{(n + 3) (n + 6)} + . . . . . . + \frac{1}{6 n}

, then

lim n \to \infty S_{n}

Q. If

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

, then

\sum_{r = 1}^{n} \frac{1}{T}

is equal to:

Q. If

S_{n} = \sum_{n}^{r = 0} \frac{1}{^{n} C_{r}}

and

t_{n} = \sum_{n}^{r = 0} \frac{r}{^{n} C_{r}}

, then

\frac{t_{n}}{S_{n}}

is equal to

Q. If

S_{n} = n \sum r = 0 \frac{1}{^{n} C_{r}}

and

P_{n} = n \sum r = 0 \frac{r}{^{n} C_{r}}

then

\frac{S_{n}}{P_{n}}

Join BYJU'S Learning Program

If Sn=n∑r=1r−1∑t=0(16n nCr rCt 4t), then the value of l, where l=∞∑n=1(1−Sn) is

Sn=n∑r=1r−1∑t=0(16n nCr rCt 4t) ⇒Sn=n∑r=1(16n nCr(r−1∑t=0 rCt 4t)) =n∑r=1(16n nCr[(1+4)r−4r]) (Using (1+x)n=n∑r=0 nCrxr) =16nn∑r=1( nCr5r− nCr4r) =16n[(1+5)n−1−((1+4)n−1)] =6n−5n6n=1−(56)n Now, ∞∑n=1(1−Sn)=∞∑n=1(56)n =5/61−5/6=5

If $S_{n} = n \sum r = 1 r - 1 \sum t = 0 (\frac{1}{6^{n}}^{n} C_{r}^{r} C_{t} 4^{t}),$ then the value of $l,$ where $l = \infty \sum n = 1 (1 - S_{n})$ is