Advertisements
Advertisements
प्रश्न
Simplify the following:
`sum_("n" = 1)^10 "i"^("n" + 50)`
बेरीज
उत्तर
`sum_("n" = 1)^10 "i"^("n" + 50) = sum_("n" = 1)^10 "i"^"n" * "i"^50`
= `sum_("n" = 1)^10 "i"^"n" * "i"^48 * "i"^2`
= `- 1[sum_("n" = 1)^10 "i"^"n"]`
= `- 1[("i" + "i"^2 + "i"^3 + "i"^4) + ("i"^5 + "i"^6 + "i"^7 + "i"^8) + "i"^9 + "i"^10]`
= `- 1[0 + 0 "i" + "i"^2]`
= `- 1("i" - 1)`
= 1 – i
shaalaa.com
Introduction to Complex Numbers
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
Simplify the following:
i1947 + i1950
Simplify the following:
i1948 – i–1869
Simplify the following:
`sum_("n" = 1)^12 "i"^"n"`
Simplify the following:
`"i"^59 + 1/"i"^59`
Simplify the following:
i i2 i3 ... i2000
Choose the correct alternative:
in + in+1+ in+2 + in+3 is
Choose the correct alternative:
The value of `sum_("n" = 1)^13 ("i"^"n" + "i"^("n" - 1))` is
