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Sum of n terms of series12+16+24+40+...will be
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a
22n−1+8n
b
22n−1+6n
c
32n−1+8n
d
42n−1+8n
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Correct option is D
Let Sn=12+16+24+…+Tn Sn=12+16 +…+TnOn subtraction0=12+4+8+16+…−Tn⇒ Tn=12+42n−1−12−1=2n+1+8Sn=ΣTn=22+23+24+…+8n=222n−12−1+8n=42n−1+8nSimilar Questions
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