If a 1- a 2- a 3- - a n are in A-P- with s n as the sum of first -n- terms S 0-0- then - k -0 n n C k S k is equal toA- 2-n a1-sn-B- 2n-a1-sn-C- 2n-1-a1-sn-D- 2n-2-n a1-sn-

∑_k=0^n^nC_k S_k=∑_k=0^n^nC_k k/n[2a+(k 1)d ] =[(a_1 d/2)∑_k=0^nk^nc_k+d/2∑_k=0^nk^2 ^nC_k] = ( a_1 d/2 )n. 2^n 1+d/2[n. 2^n 1+n(n 1)2^n 2] =a_1.n.2^n 1+dn(n 1 ...

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

Byju's Answer

Standard XII

Mathematics

Inductive Step

If a 1, a 2, ...

Question

If $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ are in A.P. with $s_{n}$ as the sum of first 'n' terms $(S_{0} = 0), t h e n \sum_{k = 0}^{n}^{n} C_{k} S_{k}$ is equal to

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!

Open in App

Solution

The correct option is A
$\sum_{k = 0}^{n}^{n} C_{k} S_{k} = \sum_{k = 0}^{n}^{n} C_{k} \frac{k}{n} [2 a + (k - 1) d]$
$= [(a_{1} - \frac{d}{2}) \sum_{k = 0}^{n} k^{n} c_{k} + \frac{d}{2} \sum_{k = 0}^{n} k^{2}^{n} C_{k}]$
$= (a_{1} - \frac{d}{2}) n . 2^{n - 1} + \frac{d}{2} [n . 2^{n - 1} + n (n - 1) 2^{n - 2}]$
$= a_{1} . n {.2}^{n - 1} + d n (n - 1) 2^{n - 3}$
$= n {.2}^{n - 3} [4 a_{1} + a_{n} - a_{1}] = n {.2}^{n - 3} [3 a_{1} + a_{n}]$
$= 2^{n - 3} [2 n a_{1} + 2 n (\frac{a_{1} + a_{n}}{2})]$
$= 2^{n - 2} [n a_{1} + s_{n}]$

Suggest Corrections

Similar questions

Q. If sum of first

n

terms of an A.P. is

S_{n} = 3 n^{2} - 2 n,

then the value of

\infty \sum n = 1 (\frac{21}{(S_{n} S_{n + 2} + S_{n - 1} S_{n + 1}) - (S_{n} S_{n + 1} + S_{n - 1} S_{n + 2})})

is equal to

Q. Suppose

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P and

S_{k}

denotes the sum of first

k

terms, If

\frac{S_{n}}{S_{m}} = \frac{n^{3}}{m^{3}}

, then the ratio of

(m + 1)^{th}

term to

(n + 1)^{th}

term is

Q. If S_n =

\sum_{r = 1}^{n} \frac{1 + 2 + 2^{2} + . . . Sum to r terms}{2^{r}}

, then S_n is equal to
(a) 2ⁿ − n − 1

(b)

1 - \frac{1}{2^{n}}

(c)

n - 1 + \frac{1}{2^{n}}

(d) 2ⁿ − 1

Q. In an A.P. a₁, a₂, a₃, ..., a_n, ...., if a₁ = 1, a_n= 20 and S_n= 399, then the value of n is __________.

Q. If

S_{n} = n P + \frac{n}{2} (n - 1) Q,

where

S_{n}

denotes the sum of the first

n

terms of an A.P., then the common difference is

Join BYJU'S Learning Program

If a1,a2,a3,...,an are in A.P. with sn as the sum of first 'n' terms (S0=0), then ∑nk=0nCkSk is equal to

The correct option is A ∑nk=0nCkSk=∑nk=0nCkkn[2a+(k−1)d] =[(a1−d2)∑nk=0knck+d2∑nk=0k2 nCk] =(a1−d2)n.2n−1+d2[n.2n−1+n(n−1)2n−2] =a1.n.2n−1+dn(n−1)2n−3 =n.2n−3[4a1+an−a1]=n.2n−3[3a1+an] =2n−3[2na1+2n(a1+an2)] =2n−2[na1+sn]

If $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ are in A.P. with $s_{n}$ as the sum of first 'n' terms $(S_{0} = 0), t h e n \sum_{k = 0}^{n}^{n} C_{k} S_{k}$ is equal to