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Question
Write \[\sum^m_{r = 0} \ ^{n + r}{}{C}_r\] in the simplified form.
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Solution
We know:
\[ \because \ ^{n}{}{C}_0 = \ ^{n + 1}{}{C}_0 \]
\[ \therefore \sum^m_{r = 0} \ ^{n + r}{}{C}_r = \ ^{n + 1}{}{C}_0 + \ ^{n + 1}{}{C}_1 + \ ^{n + 2}{}{C}_2 + \ ^{n + 3}{}{C}_3 + . . . + \ ^{n + m}{}{C}_m \]
\[Using \ ^{n}{}{C}_{r - 1} + \ ^{n}{}{C}_r = \ ^{n + 1}{}{C}_r : \]
\[ \Rightarrow \sum^m_{r = 0} \ ^{n + r}{}{C}_r = \ ^{n + 2}{}{C}_1 + \ ^{n + 2}{}{C}_2 + \ ^{n + 3}{}{C}_3 + . . . + \ ^{n + m}{}{C}_m \]
\[ \Rightarrow \sum^m_{r = 0} \ ^{n + r}{}{C}_r = \ ^{n + 3}{}{C}_2 + \ ^ {n + 3}{}{C}_3 + . . . + \ ^{n + m}{}{C}_m\]
\[ \Rightarrow \sum^m_{r = 0} \ ^{n + r}{}{C}_r = \ ^{n + m + 1}{}{C}_m\]
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