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# hey guys, i have a quesiton on how to add these binomials

0
345
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hey guys, i have a quesiton on how to add these binomials

99c3 + 99c5 + 99c7 + ... + 99c97

im decently sure there is a formula for this, but i cannot remember it. help? ty

Jun 28, 2021

#1
0

Using generating functions, the sum is 2^99.

Jun 28, 2021
#2
+118118
+1

99C0+99C99=1+1=2

99C1+99C98=99+99 = 198

99C2+99C97=4851+4851=9792

If I add all those together I get   9900

So the sum is  2^99 - 9900

I think it is anyway.

Jun 28, 2021
#3
+26329
+3

$$^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97}$$

$$\small{ \begin{array}{|rcll|} \hline ^{99}C_0+~^{99}C_1+~^{99}C_2+~^{99}C_3 + \ldots + ~^{99}C_{97}+~^{99}C_{98}+~^{99}C_{99} &=& 2^{99} \\\\ ~^{99}C_1+~^{99}C_3+~^{99}C_5 + \ldots + ~^{99}C_{97}+~^{99}C_{99} &=& \dfrac{2^{99}}{2} \\\\ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} &=& \dfrac{2^{99}}{2} -~^{99}C_1-~^{99}C_{99} \\\\ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} &=& 2^{98} -99-1 \\\\ \mathbf{ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} } &=& \mathbf{2^{98} -100 } \\ \hline \end{array} }$$

Jun 28, 2021
edited by heureka  Jun 28, 2021
#4
+118118
0

Thanks Heureka,

I did not read the question properly.   I didn't notice that it was only odd values of r.

Melody  Jun 28, 2021
#5
+367
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thanks heureka!