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Question

If Cr=Cr25 and C0+5C1+9C2++101·C25=225·k then k is equal to


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Solution

Step 1. Find the value of k:

Let S=C025+5C125+9C225+.+97C2425+101C2525=225k …..(1)

Reverse the series and apply Cnn=Cn-rn in all coefficients,

S=101C025+97C125++5C2425+C2525 …..(2)

Step 2. Add equation (1) and (2),we get

2S=102[C025+C125++C525]

S=51×225

k=51

Hence, The value of k=51


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