1
Question
If n>2 then sum the series
n∑r=2(−2)r∣∣
∣∣n−2Cr−2n−2Cr−1n−2Cr−3112−10∣∣
∣∣
n∑r=2(−2)r∣∣ ∣∣n−2Cr−2n−2Cr−1n−2Cr−3112−10∣∣ ∣∣
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Solution
Applying C1+C2+2C2 thus making two zeros in C1
Also n−2Cr−2+n−2Cr+2.n−2Cr−1
=[n−2Cr−2+n−2Cr−1]+[n−2Cr−1+n−2Cr]
=n−1Cr−1+n−1Cr=nCr
∴Δ=∑nr=2(−2)rnCr
=C2(−2)2+C3(−2)3+C4(−2)4++Cn(−2)n
=(1+x)n−[C0+C1x], where x=−2
=(1−2)n−[1+n(−2)]=2n−1+(−1)n.
Also n−2Cr−2+n−2Cr+2.n−2Cr−1
=[n−2Cr−2+n−2Cr−1]+[n−2Cr−1+n−2Cr]
=n−1Cr−1+n−1Cr=nCr
∴Δ=∑nr=2(−2)rnCr
=C2(−2)2+C3(−2)3+C4(−2)4++Cn(−2)n
=(1+x)n−[C0+C1x], where x=−2
=(1−2)n−[1+n(−2)]=2n−1+(−1)n.
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