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Particular Solution of a Differential Equation

If 50 C 0+51 ...

Question

If $^{50} C_{0} +^{51} C_{1} +^{52} C_{2} + . . . . . +^{100} C_{50} =^{n} C_{k}$ , then the value of $n + k$ is

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Solution

Let
$S =^{50} C_{0} +^{51} C_{1} +^{52} C_{2} + . . . . . +^{100} C_{50}$
$=^{51} C_{0} +^{51} C_{1} +^{52} C_{2} + . . . . . +^{100} C_{50}$

We know $^{N} C_{r} +^{N} C_{r - 1} =^{N + 1} C_{r}$

$=^{52} C_{1} +^{52} C_{2} +^{53} C_{3} + . . . . . +^{100} C_{50}$
$. . .$
$. . .$
$. . .$
$=^{100} C_{49} +^{100} C_{50} =^{101} C_{50} =^{101} C_{51}$

Hence the value of $n + k = 151$ or $n + k = 152$

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