binom{10}{1} + binom{10}{2} + binom{11}{3} + | vedclass

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(D) We use the Pascal's identity: $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{10}{1} + \binom{10}{2} + \binom{11}{3} +

Vedclass · Accepted Answer

$\binom{14}{5}$

Vedclass · Answer

(D) We use the Pascal's identity: $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{10}{1} + \binom{10}{2} + \binom{11}{3} + \binom{12}{4} + \binom{13}{5}$.Step $1$: $\binom{10}{1} + \binom{10}{2} = \binom{11}{2}$.Step $2$: $\binom{11}{2} + \binom{11}{3} = \binom{12}{3}$.Step $3$: $\binom{12}{3} + \binom{12}{4} = \binom{13}{4}$.Step $4$: $\binom{13}{4} + \binom{13}{5} = \binom{14}{5}$.Thus,the final result is $\binom{14}{5}$.

$\binom{10}{1} + \binom{10}{2} + \binom{11}{3} + \binom{12}{4} + \binom{13}{5} = \dots$

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