(4-1/n)+(4-2/n)+(4-3/n)+... के n पदों का योग ज्ञात कीजिए
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Step-by-step explanation:
inshallah it helps uuh ...❤...
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Step-by-step explanation:
a=4-1/n
d=(4-2/n)-(4-1/n)=-1/n
Sn=n/2 [2a+(n-1) d]
=n/2 [2 (4-1/n)+(n-1)(-1/n)]
=n/2[8-2/n+(-1+1/n)]
=n/2 [8n-2-n+1]
=1/2 (7n-1)
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