How do you use the Maclaurin series for # f(x) = (2+x)^5#?

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Answer 1 · 2016-09-22T05:39:46

#f(x)=32+80x+80x^2+40x^3+10x^4+x^5#

According to Maclaurin series,

#f(x)=f(0)+f'(0)x+f''(0)x^2/(2!)x^2+f'''(0)x^3/(3!)+f''''(0)x^4/(4!)+....#

As #f(x)=(2+x)^5#, #f(0)=32#

#f'(x)=5(2+x)^4# and #f''(0)=80#

#f''(x)=5xx4(2+x)^3# and #f'''(0)=160#

#f'''(x)=5xx4xx3(2+x)^2# and #f''''(0)=240#

#f''''(x)=5xx4xx3xx2(2+x)# and #f''''(0)=240#

#f'''''(x)=5xx4xx3xx2xx1# and #f''''(0)=120# and #f''''''(x)=0#

Hence, #f(x)=32+80x+160x^2/2+240x^3/6+240x^4/24+120x^5/120#

or #f(x)=32+80x+80x^2+40x^3+10x^4+x^5#