Verified Answer :

apologiabiology

the nth term in the expansion (a+b)^m=
(\(\frac{(m)!}{(n-1)!(m-(n-1))!}\))a^(m-(n-1))b^(n-1)

so

8th term
(\(\frac{(14)!}{(8-1)!(14-(8-1))!}\))a^(14-(8-1))b^(8-1)=
(\(\frac{(14)!}{7!(7)!}\))a^(7)b^(7)=
(\(\frac{14*13*12*11*10*9*8}{7!}\))a^(7)b^(7)=
3432a⁷b⁷ is answer

konrad509

\((a+b)^n\\\displaystyle T_r=\binom{n}{r-1}a^{n-r+1}b^{r-1}\\\\T_8=\binom{14}{8-1}a^{14-8+1}\cdot b^{8-1}\\T_8=\binom{14}{7}a^7b^7\\T_8=\dfrac{14!}{7!7!}\cdota^7b^7\\T_8=\dfrac{8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14}{2\cdot3\cdot4\cdot5\cdot6\cdot7}\cdot a^7b^7\\T_8=3432a^7b^7\)

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