Question

The ${49}^{th}$ term in the expansion of
$(a+b{)}^{100}$ is

• A. ${}^{100}{C}_{48}{a}^{52}{b}^{48}$
• B. ${}^{100}{C}_{49}{a}^{49}{b}^{51}$
• C. ${}^{100}{C}_{48}{a}^{48}{b}^{52}$
• D. ${}^{100}{C}_{49}{a}^{51}{b}^{49}$

Answer 1

The correct option is A ${}^{100}{C}_{48}{a}^{52}{b}^{48}$

The expansion of $(a+b{)}^{100}$ is
${}^{100}{C}_0{a}^{100}{b}^0+{}^{100}{C}_1{a}^{99}{b}^1+{}^{100}{C}_2{a}^{98}{b}^2+\cdot \cdot \cdot +{}^{100}{C}_{100}{a}^0{b}^{100}$
We observe that,
$\to$Sum of the power of $a$ and $b$ should be equal to index i.e $100$
$\to$ The first term is ${}^{100}{C}_0{a}^{100}{b}^0$
Second term is ${}^{100}{C}_1{a}^{99}{b}^1$
Third term is ${}^{100}{C}_2{a}^{98}{b}^2$,
Similarly
Fourty ninth term is ${}^{100}{C}_{48}{a}^{52}{b}^{48}$

Hence, the correct answer is Option d.