If total number of runs scored in n matches is n-142n-1-n-2 where n - 1- and the runs scored in the kth match are given by k-2n-1-k- where 1- k - n- Find n

S_n=∑_k=1^nk.2^n+1 k=2^n+1∑_k=1^nk.2^ k =2^n+1[12(1 12^n)1 12]11 12=2^2n+2[1 12^n n2^n+1]=2^n+2 4 2n=2(2^n+1 2 n) so, S_n=n+14(2^n+1 2 n) (as gi ...

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

Byju's Answer

Standard XII

Mathematics

Combination of r Things from n Things When All Are Not Different

If total numb...

Question

If total number of runs scored in n matches is $(\frac{n + 1}{4}) (2^{n + 1} - n - 2)$ where $n > 1,$ and the runs scored in the $k^{t h}$ match are given by $k {.2}^{n + 1 - k}$ , where $1 \leq k \leq n$ . Find $n$

Open in App

Solution

$S_{n} = \sum_{k = 1}^{n} k {.2}^{n + 1 - k} = 2^{n + 1} \sum_{k = 1}^{n} k {.2}^{- k}$
$= 2^{n + 1} [\frac{\frac{1}{2} (1 - \frac{1}{2^{n}})}{1 - \frac{1}{2}}] \frac{1}{1 - \frac{1}{2}} = 2^{2 n + 2} [1 - \frac{1}{2^{n}} - \frac{n}{2^{n + 1}}] = 2^{n + 2} - 4 - 2 n = 2 (2^{n + 1} - 2 - n)$
so, $S_{n} = \frac{n + 1}{4} (2^{n + 1} - 2 - n)$ (as given)
So, $\frac{n + 1}{4} (2^{n + 1} - 2 - n) = 2 (2^{n - 1} - n - 2)$
$n + 1 = 8, n = 7$

Suggest Corrections

Similar questions

Q. If total number of runs scored in

n

matches is

(\frac{n + 1}{4}) (2^{n + 1} - n - 2)

where

n > 1,

and the runs scored in the

k^{th}

match are given by

k \cdot 2^{n + 1 - k}

where

1 \leq k \leq n,

then the value of

n

Q. If total number of runs scored in n matches is

(\frac{n + 1}{4}) (2^{n + 1} - n - 2)

where n > 1, and the runs scored
in the

k^{t h}

match are given by k

2^{n + 1}

where

1 < k < n

. Find n.

Q. If total number of runs scored in

n

matches is

(\frac{n + 1}{4}) (2^{n + 1} - n - 2)

where

n > 1,

and the runs scored in the

k^{th}

match are given by

k \cdot 2^{n + 1 - k}

where

1 \leq k \leq n,

then the value of

n

Q. A cricketer has to score

4500

run. Let

a_{n}

denotes the number of run he scores in the

n^{t h}

match. If

a_{1} = a_{2} = . . . . . . = a_{10} = 150

and

a_{10}, a_{11}, a_{12}

,... are in A.P. with common difference -2, then find the total number of matches played by him to score 4500 runs

If $D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $n \sum k = 1 D_{k} = 56,$ where $n \in N,$ then $n$ is equal to

Join BYJU'S Learning Program

If total number of runs scored in n matches is (n+14)(2n+1−n−2) where n>1, and the runs scored in the kth match are given by k.2n+1−k, where 1≤k≤n. Find n

Sn=∑nk=1k.2n+1−k=2n+1∑nk=1k.2−k=2n+1[12(1−12n)1−12]11−12=22n+2[1−12n−n2n+1]=2n+2−4−2n=2(2n+1−2−n)so, Sn=n+14(2n+1−2−n) (as given)So, n+14(2n+1−2−n)=2(2n−1−n−2)n+1=8,n=7

If total number of runs scored in n matches is $(\frac{n + 1}{4}) (2^{n + 1} - n - 2)$ where $n > 1,$ and the runs scored in the $k^{t h}$ match are given by $k {.2}^{n + 1 - k}$ , where $1 \leq k \leq n$ . Find $n$