Question

Find the coefficient of x5 in the expansion of (1 + x)21 + (1 + x)22 + ... + (1 + x) 30​

Answer 1

Answer:

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Step-by-step explanation:

S=(1+x)

21

+(1+x)

22

+(1+x)

23

+...(1+x)

30

...(i)

(1+x)S=(1+x)

22

+(1+x)

23

...(1+x)

30

+(1+x)

31

...(ii)

Subtracting (i) from (ii), we get

xS=(1+x)

31

−(1+x)

21

S=

x

(1+x)

31

−

x

(1+x)

21

Coefficient of x

r

=

31

C

r

x

r−1

−

21

C

r

x

r−1

=(

31

C

r

−

21

C

r

)x

r−1

... (a)

Therefore the coefficient of x

5

implies

r−1=5

r=6

Substituting in (a), we get coefficient of x

5

=(

31

C

6

−

21

C

6

)