Question

please solve this question ..........please......​

Answer 1

$\huge\underline{\overline{\mid{\bold{\red{ANSWER-}}\mid}}}$

To Prove ∑nr=03r×Crn=4n by binomial expansion

(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=

∑nr=0Crn×an−r×br

if

we compare the two terms

∑nr=03r×Crn=∑nr=0Crn×an−r×br

hence if we make a=1 and b=3

we get ∑nr=0Crn×1n−r×3r

which is same as given term,

∑nr=03r×Crn

putting the same values in (a+b)n=(1+3)n=4n

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\blacksquare \:$

FORMULA TO BE IMPLEMENTED

We are aware of the Binomial Expansion that

${(x + y)}^{n} = \: ^{n}C_0 \: {x}^{n} + \: ^{n}C_1 \: {x}^{n - 1} y + \: ^{n}C_2 \: {x}^{n - 2} {y}^{2} + \: ^{n}C_3 \: {x}^{n - 3} {y}^{3} + ........ + \: ^{n}C_n \: {y}^{n}$

$\blacksquare$

CALCULATION

Here

${(x + y)}^{n} = \: ^{n}C_0 \: {x}^{n} + \: ^{n}C_1 \: {x}^{n - 1} y + \: ^{n}C_2 \: {x}^{n - 2} {y}^{2} + \: ^{n}C_3 \: {x}^{n - 3} {y}^{3} + ........ + \: ^{n}C_n \: {y}^{n}$

Putting x = 1 & y = 3 we get

${(1 + 3)}^{n} = \: ^{n}C_0 \: {1}^{n} + \: ^{n}C_1 \: {1}^{n - 1} \times 3 \: + \: ^{n}C_2 \: {1}^{n - 2} \times {3}^{2} + \: ^{n}C_3 \: {1}^{n - 3} \times {3}^{3} + ........ + \: ^{n}C_n \: \times {3}^{n}$

$\implies \: \displaystyle \: {(4)}^{n} = \: ^{n}C_0 \: + \: ^{n}C_1 \: \times 3 \: + \: ^{n}C_2 \: \times {3}^{2} + \: ^{n}C_3 \: \times {3}^{3} + ........ + \: ^{n}C_n \: \times {3}^{n}$

$\implies \: \displaystyle \: \: ^{n}C_0 \: + \: ^{n}C_1 \: \times 3 \: + \: ^{n}C_2 \: \times {3}^{2} + \: ^{n}C_3 \: \times {3}^{3} + ........ + \: ^{n}C_n \: \times {3}^{n} = {4}^{n}$

$\implies \: \sum_{r=0}^{n} 3^{r} \: \: ^{n}C_r = {4}^{n}$